description: <image>
  
  William has two arrays a and b, each consisting of n items.
  
  For some segments l..r of these arrays William wants to know if it is possible to equalize the values of items in these segments using a balancing operation. Formally, the values are equalized if for each i from l to r holds a_i = b_i.
  
  To perform a balancing operation an even number of indices must be selected, such that l ≤ pos_1 < pos_2 < ... < pos_k ≤ r. Next the items of array a at positions pos_1, pos_3, pos_5, ... get incremented by one and the items of array b at positions pos_2, pos_4, pos_6, ... get incremented by one.
  
  William wants to find out if it is possible to equalize the values of elements in two arrays for each segment using some number of balancing operations, and what is the minimal number of operations required for that. Note that for each segment the operations are performed independently.
  
  Input
  
  The first line contains a two integers n and q (2 ≤ n ≤ 10^5, 1 ≤ q ≤ 10^5), the size of arrays a and b and the number of segments.
  
  The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^9).
  
  The third line contains n integers b_1, b_2, ..., b_n (0 ≤ b_i ≤ 10^9).
  
  Each of the next q lines contains two integers l_i and r_i (1 ≤ l_i < r_i ≤ n), the edges of segments.
  
  Output
  
  For each segment output a single number — the minimal number of balancing operations needed or "-1" if it is impossible to equalize segments of arrays.
  
  Example
  
  Input
  
  
  8 5
  0 1 2 9 3 2 7 5
  2 2 1 9 4 1 5 8
  2 6
  1 7
  2 4
  7 8
  5 8
  
  
  Output
  
  
  1
  3
  1
  -1
  -1
  
  Note
  
  For the first segment from 2 to 6 you can do one operation with pos = [2, 3, 5, 6], after this operation the arrays will be: a = [0, 2, 2, 9, 4, 2, 7, 5], b = [2, 2, 2, 9, 4, 2, 5, 8]. Arrays are equal on a segment from 2 to 6 after this operation.
  
  For the second segment from 1 to 7 you can do three following operations: 
  
    1. pos = [1, 3, 5, 6] 
    2. pos = [1, 7] 
    3. pos = [2, 7] 
  
  
  
  After these operations, the arrays will be: a = [2, 2, 2, 9, 4, 2, 7, 5], b = [2, 2, 2, 9, 4, 2, 7, 8]. Arrays are equal on a segment from 1 to 7 after these operations.
  
  For the third segment from 2 to 4 you can do one operation with pos = [2, 3], after the operation arrays will be: a = [0, 2, 2, 9, 3, 2, 7, 5], b = [2, 2, 2, 9, 4, 1, 5, 8]. Arrays are equal on a segment from 2 to 4 after this operation.
  
  It is impossible to equalize the fourth and the fifth segment.
solution: #include <bits/stdc++.h>
  using namespace std;
  mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
  mt19937 rnf(2106);
  const int N = 100005;
  int n, qq;
  int a[N];
  struct ban {
    long long s;
    long long maxu;
    long long minu;
    ban() { s = maxu = minu = 0; }
    ban(int x) {
      s = x;
      maxu = max(0, x);
      minu = min(0, x);
    }
  };
  ban t[N * 4];
  ban mer(const ban& l, const ban& r) {
    ban res;
    res.s = l.s + r.s;
    res.maxu = max(l.maxu, l.s + r.maxu);
    res.minu = min(l.minu, l.s + r.minu);
    return res;
  }
  void bil(int tl, int tr, int pos) {
    if (tl == tr) {
      t[pos] = ban(a[tl]);
      return;
    }
    int m = (tl + tr) / 2;
    bil(tl, m, pos * 2);
    bil(m + 1, tr, pos * 2 + 1);
    t[pos] = mer(t[pos * 2], t[pos * 2 + 1]);
  }
  ban qry(int tl, int tr, int l, int r, int pos) {
    if (l > r) return ban();
    if (tl == l && tr == r) return t[pos];
    int m = (tl + tr) / 2;
    return mer(qry(tl, m, l, min(m, r), pos * 2),
               qry(m + 1, tr, max(m + 1, l), r, pos * 2 + 1));
  }
  void solv() {
    cin >> n >> qq;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) {
      int x;
      cin >> x;
      a[i] -= x;
    }
    bil(1, n, 1);
    while (qq--) {
      int l, r;
      cin >> l >> r;
      ban u = qry(1, n, l, r, 1);
      if (u.maxu == 0 && u.s == 0) {
        cout << -u.minu << "\n";
      } else
        cout << "-1\n";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(0);
    int tt = 1;
    while (tt--) {
      solv();
    }
    return 0;
  }
  