Problem: It turns out that you are a great fan of rock band AC/PE. Peter learned that and started the following game: he plays the first song of the list of n songs of the group, and you have to find out the name of the song. After you tell the song name, Peter immediately plays the following song in order, and so on.

The i-th song of AC/PE has its recognizability pi. This means that if the song has not yet been recognized by you, you listen to it for exactly one more second and with probability of pi percent you recognize it and tell it's name. Otherwise you continue listening it. Note that you can only try to guess it only when it is integer number of seconds after the moment the song starts playing.

In all AC/PE songs the first words of chorus are the same as the title, so when you've heard the first ti seconds of i-th song and its chorus starts, you immediately guess its name for sure.

For example, in the song Highway To Red the chorus sounds pretty late, but the song has high recognizability. In the song Back In Blue, on the other hand, the words from the title sound close to the beginning of the song, but it's hard to name it before hearing those words. You can name both of these songs during a few more first seconds.

Determine the expected number songs of you will recognize if the game lasts for exactly T seconds (i. e. you can make the last guess on the second T, after that the game stops).

If all songs are recognized faster than in T seconds, the game stops after the last song is recognized.

Input

The first line of the input contains numbers n and T (1 ≤ n ≤ 5000, 1 ≤ T ≤ 5000), separated by a space. Next n lines contain pairs of numbers pi and ti (0 ≤ pi ≤ 100, 1 ≤ ti ≤ T). The songs are given in the same order as in Petya's list.

Output

Output a single number — the expected number of the number of songs you will recognize in T seconds. Your answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Examples

Input

2 2
50 2
10 1


Output

1.500000000


Input

2 2
0 2
100 2


Output

1.000000000


Input

3 3
50 3
50 2
25 2


Output

1.687500000


Input

2 2
0 2
0 2


Output

1.000000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/stack:20000000")
using namespace std;
double __begin;
template <typename T1, typename T2, typename T3>
struct triple {
  T1 a;
  T2 b;
  T3 c;
  triple(){};
  triple(T1 _a, T2 _b, T3 _c) : a(_a), b(_b), c(_c) {}
};
template <typename T1, typename T2, typename T3>
bool operator<(const triple<T1, T2, T3> &t1, const triple<T1, T2, T3> &t2) {
  if (t1.a != t2.a)
    return t1.a < t2.a;
  else
    return t1.b < t2.b;
}
template <typename T1, typename T2, typename T3>
inline std::ostream &operator<<(std::ostream &os, const triple<T1, T2, T3> &t) {
  return os << "(" << t.a << ", " << t.b << ", " << t.c << ")";
}
inline int bits_count(int v) {
  v = v - ((v >> 1) & 0x55555555);
  v = (v & 0x33333333) + ((v >> 2) & 0x33333333);
  return ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24;
}
inline int bits_count(long long v) {
  int t = v >> 32;
  int p = (v & ((1LL << 32) - 1));
  return bits_count(t) + bits_count(p);
}
unsigned int reverse_bits(register unsigned int x) {
  x = (((x & 0xaaaaaaaa) >> 1) | ((x & 0x55555555) << 1));
  x = (((x & 0xcccccccc) >> 2) | ((x & 0x33333333) << 2));
  x = (((x & 0xf0f0f0f0) >> 4) | ((x & 0x0f0f0f0f) << 4));
  x = (((x & 0xff00ff00) >> 8) | ((x & 0x00ff00ff) << 8));
  return ((x >> 16) | (x << 16));
}
inline int sign(int x) { return (x >> 31) | (-x >> 31); }
inline bool ispow2(int x) { return (x != 0 && (x & (x - 1)) == 0); }
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::pair<T1, T2> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
  bool first = true;
  os << "[";
  for (unsigned int i = 0; i < v.size(); i++) {
    if (!first) os << ", ";
    os << v[i];
    first = false;
  }
  return os << "]";
}
template <typename T>
inline std::ostream &operator<<(std::ostream &os, const std::set<T> &v) {
  bool first = true;
  os << "[";
  for (typename std::set<T>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T1, typename T2>
inline std::ostream &operator<<(std::ostream &os, const std::map<T1, T2> &v) {
  bool first = true;
  os << "[";
  for (typename std::map<T1, T2>::const_iterator ii = v.begin(); ii != v.end();
       ++ii) {
    if (!first) os << ", ";
    os << *ii;
    first = false;
  }
  return os << "]";
}
template <typename T, typename T2>
void printarray(T a[], T2 sz, T2 beg = 0) {
  for (T2 i = beg; i < sz; i++) cout << a[i] << " ";
  cout << endl;
}
inline long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n & 1) res *= x;
    x *= x;
    n >>= 1;
  }
  return res;
}
inline long long powmod(long long x, long long n, long long _mod) {
  long long res = 1;
  while (n) {
    if (n & 1) res = (res * x) % _mod;
    x = (x * x) % _mod;
    n >>= 1;
  }
  return res;
}
inline long long gcd(long long a, long long b) {
  long long t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline int gcd(int a, int b) {
  int t;
  while (b) {
    a = a % b;
    t = a;
    a = b;
    b = t;
  }
  return a;
}
inline long long lcm(int a, int b) { return a / gcd(a, b) * (long long)b; }
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long gcd(long long a, long long b, long long c) {
  return gcd(gcd(a, b), c);
}
inline int gcd(int a, int b, int c) { return gcd(gcd(a, b), c); }
inline long long lcm(long long a, long long b, long long c) {
  return lcm(lcm(a, b), c);
}
inline long long lcm(int a, int b, int c) {
  return lcm(lcm(a, b), (long long)c);
}
inline long long max(long long a, long long b) { return (a > b) ? a : b; }
inline int max(int a, int b) { return (a > b) ? a : b; }
inline double max(double a, double b) { return (a > b) ? a : b; }
inline long long max(long long a, long long b, long long c) {
  return max(a, max(b, c));
}
inline int max(int a, int b, int c) { return max(a, max(b, c)); }
inline double max(double a, double b, double c) { return max(a, max(b, c)); }
inline long long min(long long a, long long b) { return (a < b) ? a : b; }
inline int min(int a, int b) { return (a < b) ? a : b; }
inline double min(double a, double b) { return (a < b) ? a : b; }
inline long long min(long long a, long long b, long long c) {
  return min(a, min(b, c));
}
inline int min(int a, int b, int c) { return min(a, min(b, c)); }
inline double min(double a, double b, double c) { return min(a, min(b, c)); }
template <class T>
inline void getar(T a, int n, int m) {
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; ++j) {
      scanf("%d", &a[i][j]);
    }
}
inline void getar(int *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%d", a + ii);
  }
}
inline void getar(pair<int, int> *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%d%d", &a[ii].first, &a[ii].second);
  }
}
inline void getar(long long *a, int n) {
  for (int ii = 0; ii < n; ii++) {
    scanf("%I64d", a + ii);
  }
}
template <class T>
inline void cinarr(T a, int n) {
  for (int i = 0; i < n; ++i) cin >> a[i];
}
template <class T>
inline bool scan_d(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return 1;
}
inline void out(int x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int n, m, t[5005], a, flag;
double dp[2][5005], p[5005], ans;
int main() {
  while (scanf("%d%d", &n, &m) != EOF) {
    flag = 0;
    for (int j = 0; j < m + 1; j++) {
      dp[0][j] = dp[1][j] = 0;
    }
    for (int i = 1; i < n + 1; i++) {
      scanf("%d%d", &a, &t[i]);
      p[i] = 1.0 - (double)a / 100.0;
    }
    dp[flag][0] = 1.0;
    ans = 0;
    for (int i = 1; i < n + 1; i++) {
      double po = 1.0;
      po = pow(1.0 * (p[i]), t[i] - 1);
      int now = flag ^ 1, last = flag;
      for (int j = 1; j < m + 1; j++) {
        double s = dp[now][j - 1];
        if (j - 1 >= t[i]) s -= dp[last][j - 1 - t[i]] * po;
        s *= p[i];
        if (j >= t[i]) {
          s += dp[last][j - t[i]] * po * p[i];
        }
        s += dp[last][j - 1] * (1.0 - p[i]);
        dp[now][j] = s;
        ans += dp[now][j];
      }
      dp[flag][0] = 0;
      flag ^= 1;
    }
    printf("%.10f\n", ans);
  }
  return 0;
}