1005_E1. Median on Segments (Permutations Edition)
You are given a permutation p_1, p_2, ..., p_n. A permutation of length n is a sequence such that each integer between 1 and n occurs exactly once in the sequence.

Find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of the median of p_l, p_{l+1}, ..., p_r is exactly the given number m.

The median of a sequence is the value of the element which is in the middle of the sequence after sorting it in non-decreasing order. If the length of the sequence is even, the left of two middle elements is used.

For example, if a=[4, 2, 7, 5] then its median is 4 since after sorting the sequence, it will look like [2, 4, 5, 7] and the left of two middle elements is equal to 4. The median of [7, 1, 2, 9, 6] equals 6 since after sorting, the value 6 will be in the middle of the sequence.

Write a program to find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of the median of p_l, p_{l+1}, ..., p_r is exactly the given number m.

Input

The first line contains integers n and m (1 ≤ n ≤ 2⋅10^5, 1 ≤ m ≤ n) — the length of the given sequence and the required value of the median.

The second line contains a permutation p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). Each integer between 1 and n occurs in p exactly once.

Output

Print the required number.

Examples

Input

5 4
2 4 5 3 1


Output

4


Input

5 5
1 2 3 4 5


Output

1


Input

15 8
1 15 2 14 3 13 4 8 12 5 11 6 10 7 9


Output

48

Note

In the first example, the suitable pairs of indices are: (1, 3), (2, 2), (2, 3) and (2, 4).
```python
n, m = map(int, input().split(' '))
aa = list(map(int, input().split(' ')))

def countMoreLess(vs, value):
    #print(vs)
    cur = 0
    res = { cur: 1 }

    for v in vs:
        cur += 1 if v > value else -1
        res[cur] = res.get(cur, 0) + 1
    #print(res)
    return res

pos = aa.index(m)

leftCounts = countMoreLess(list(reversed(aa[0:pos])), m)
rightCounts = countMoreLess(aa[pos+1:], m)

res = 0
for dif, count in leftCounts.items():
    res += count * rightCounts.get(-dif, 0)
    res += count * rightCounts.get(-dif+1, 0)

print(res)

```