#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
char s[3005], t[3005];
int dp[3005][3005];
int main() {
  int n, m, x, y;
  scanf("%d %d", &n, &m);
  scanf("%s", s + 1);
  strcpy(t + 1, s + 1);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    y++;
    sort(s + x, s + y);
    sort(t + x, t + y);
    reverse(t + x, t + y);
  }
  dp[0][0] = 1;
  int atleast = 0, atmost = 0;
  for (int i = 1; i <= n; i++) {
    atleast += s[i] == '1';
    atmost += t[i] == '1';
    for (int j = atleast; j <= atmost; j++) {
      dp[i][j] = dp[i - 1][j] + (j > 0 ? dp[i - 1][j - 1] : 0);
      if (dp[i][j] >= MOD) dp[i][j] -= MOD;
    }
  }
  printf("%d\n", dp[n][atleast]);
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

There is a string S of length N consisting of characters `0` and `1`. You will perform the following operation for each i = 1, 2, ..., m:

* Arbitrarily permute the characters within the substring of S starting at the l_i-th character from the left and extending through the r_i-th character.



Here, the sequence l_i is non-decreasing.

How many values are possible for S after the M operations, modulo 1000000007(= 10^9+7)?

Constraints

* 2≦N≦3000
* 1≦M≦3000
* S consists of characters `0` and `1`.
* The length of S equals N.
* 1≦l_i < r_i≦N
* l_i ≦ l_{i+1}

Input

The input is given from Standard Input in the following format:


N M
S
l_1 r_1
:
l_M r_M


Output

Print the number of the possible values for S after the M operations, modulo 1000000007.

Examples

Input

5 2
01001
2 4
3 5


Output

6


Input

9 3
110111110
1 4
4 6
6 9


Output

26


Input

11 6
00101000110
2 4
2 3
4 7
5 6
6 10
10 11


Output

143

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define MOD 1000000007
char s[3005], t[3005];
int dp[3005][3005];
int main() {
  int n, m, x, y;
  scanf("%d %d", &n, &m);
  scanf("%s", s + 1);
  strcpy(t + 1, s + 1);
  for (int i = 0; i < m; i++) {
    scanf("%d %d", &x, &y);
    y++;
    sort(s + x, s + y);
    sort(t + x, t + y);
    reverse(t + x, t + y);
  }
  dp[0][0] = 1;
  int atleast = 0, atmost = 0;
  for (int i = 1; i <= n; i++) {
    atleast += s[i] == '1';
    atmost += t[i] == '1';
    for (int j = atleast; j <= atmost; j++) {
      dp[i][j] = dp[i - 1][j] + (j > 0 ? dp[i - 1][j - 1] : 0);
      if (dp[i][j] >= MOD) dp[i][j] -= MOD;
    }
  }
  printf("%d\n", dp[n][atleast]);
  return 0;
}

```