#include <bits/stdc++.h>
using namespace std;
double dp[2][5001];
double prob[5001];
double pows[5001];
double EPS = 1e-20;
int times[5001];
int n;
int maxtime;
inline void compute(int k) {
  memset(pows, 0, times[k] * sizeof(double));
  pows[0] = 1;
  for (int j = 1; j < times[k]; j++) {
    pows[j] = pows[j - 1] * (1 - prob[k]);
    if (pows[j] < EPS) {
      pows[j] = 0;
      break;
    }
  }
}
inline double getupdate(int k) {
  int kk = k & 1;
  double sum = 0;
  for (int j = 1; j <= maxtime; j++) {
    if (k == n) {
      sum += dp[kk][j];
    } else if (maxtime - j < times[k]) {
      sum += dp[kk][j] * pows[maxtime - j];
    }
  }
  return sum;
}
int main() {
  cin >> n >> maxtime;
  for (int i = 0; i < n; i++) {
    int p, t;
    cin >> p >> t;
    prob[i] = p / 100.0;
    times[i] = t;
  }
  double ans = 0;
  memset(dp, 0, sizeof dp);
  compute(0);
  for (int j = 1; j <= min(times[0] - 1, maxtime); j++) {
    dp[1][j] = pows[j - 1] * prob[0];
  }
  if (times[0] <= maxtime) {
    dp[1][times[0]] = pows[times[0] - 1];
  }
  double add[maxtime + 2];
  double remove[maxtime + 2];
  for (int k = 2; k <= n; k++) {
    memset(add, 0, sizeof add);
    memset(remove, 0, sizeof remove);
    compute(k - 1);
    ans += (k - 1) * getupdate(k - 1);
    double curr = 0;
    double p = prob[k - 1];
    double q = 1 - prob[k - 1];
    int t = times[k - 1];
    int kprev = (k - 1) & 1;
    int kcurr = k & 1;
    for (int j = 1; j <= maxtime; j++) {
      dp[kcurr][j] = curr;
      curr -= remove[j];
      curr *= q;
      curr += add[j];
      if (t > 1) {
        curr += dp[kprev][j] * prob[k - 1];
        int endt = j + times[k - 1] - 1;
        if (endt <= maxtime) {
          remove[endt] += dp[kprev][j] * pows[t - 2] * p;
          add[endt] = dp[kprev][j] * pows[t - 1];
          remove[endt + 1] += add[endt];
        }
      } else {
        curr += dp[kprev][j];
        remove[j + 1] += dp[kprev][j];
      }
      if (curr < EPS) curr = 0;
    }
  }
  ans += n * getupdate(n);
  cout << setprecision(10) << fixed << ans << endl;
}
