#include <bits/stdc++.h>
//#define int long long
#define INF 0x3f3f3f3f
#define mod 1000000007
#define endl '\n'
#define rep(i, st, ed) for (int (i) = (st); (i) <= (ed);++(i))
#define pre(i, ed, st) for (int (i) = (ed); i >= (st);--(i))
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
inline int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
inline int lowbit(int x) { return x & -x; }
const int N = 1200;
int n, m;
int w[N], v[N];
int f[N];
int g[N]; // 价值为f[i] 的方案数 
void solve() {
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)cin >> w[i] >> v[i];
  // 价值为f[0] 即价值为0时的方案数为 1 那就是都不选 
  g[0] = 1;
  // 01背包
  for (int i = 1; i <= n; ++i) {
    for (int j = m; j >= w[i]; --j) {
      int maxn = max(f[j], f[j - w[i]] + v[i]);
      int cnt = 0;
      if (maxn == f[j])cnt += g[j];
      if (maxn == f[j - w[i]] + v[i])cnt += g[j - w[i]];
      f[j] = maxn;
      g[j] = cnt % mod;
    }
  }
  // 求出价值最大为多少
  int cnt = 0;
  for (int i = 0; i <= m; ++i) {
    cnt = max(cnt, f[i]);
  }
  //枚举体积 如果当前体积下价值与最大价值相等  那么答案加上 g[j]
  int res = 0;
  for (int i = 0; i <= m; ++i) {
    if (f[i] == cnt)
      res = (res + g[i]) % mod;
  }
  cout << res << endl;
}
signed main() {
  // int t;cin >> t;
  // while(t--)
  solve();
  return 0;
}