#include <bits/stdc++.h>
using namespace std;
#define REP(i, n) for (int i = 0; i < int(n); ++i)
#define REPE(i, a, b) for (int i = (a); i <= int(b); ++i)
using LL = long long;

int n, m;
char s[3010];
int sum[3010];

int dp[3010][3010];
int a[3010];

const int mod = 1000000007;
int f(int l, int r, int c) {
  if (l > r && c) return 0;
  if (l > n) return c == 0;
  if (a[l] >= r) {
    c += sum[a[l]] - sum[r];
    r = a[l];
  }
  int &ret = dp[l][c];
  if (ret >= 0) return ret;
  ret = 0;
  if (r - l + 1 - c > 0) ret += f(l + 1, r, c);
  if (c > 0) ret += f(l + 1, r, c - 1);
  if (ret >= mod) ret -= mod;
  return ret;
}

int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  for (int i = 1; i <= n; ++i)
    sum[i] = sum[i - 1] + (s[i] == '1');
  for (int i = 1; i <= n; ++i)
    a[i] = i;
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    a[l] = max(a[l], r);
  }
  memset(dp, -1, sizeof dp);
  printf("%d\n", f(1, 0, 0));
  return 0;
}


### Prompt

Please create a solution in Cpp to 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 REP(i, n) for (int i = 0; i < int(n); ++i)
#define REPE(i, a, b) for (int i = (a); i <= int(b); ++i)
using LL = long long;

int n, m;
char s[3010];
int sum[3010];

int dp[3010][3010];
int a[3010];

const int mod = 1000000007;
int f(int l, int r, int c) {
  if (l > r && c) return 0;
  if (l > n) return c == 0;
  if (a[l] >= r) {
    c += sum[a[l]] - sum[r];
    r = a[l];
  }
  int &ret = dp[l][c];
  if (ret >= 0) return ret;
  ret = 0;
  if (r - l + 1 - c > 0) ret += f(l + 1, r, c);
  if (c > 0) ret += f(l + 1, r, c - 1);
  if (ret >= mod) ret -= mod;
  return ret;
}

int main() {
  scanf("%d%d%s", &n, &m, s + 1);
  for (int i = 1; i <= n; ++i)
    sum[i] = sum[i - 1] + (s[i] == '1');
  for (int i = 1; i <= n; ++i)
    a[i] = i;
  for (int i = 0; i < m; ++i) {
    int l, r;
    scanf("%d%d", &l, &r);
    a[l] = max(a[l], r);
  }
  memset(dp, -1, sizeof dp);
  printf("%d\n", f(1, 0, 0));
  return 0;
}


```