#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 18;
const int INF = 2 * N;
struct D {
  int m[5][5];
};
D A[N * 2];
D def;
D query(int v, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return A[v];
  const int m = (l + r) / 2;
  const bool leftgo = ql <= m;
  const bool rightgo = qr > m;
  D dl, dr;
  if (leftgo) {
    dl = query(2 * v, l, m, ql, min(qr, m));
    if (!rightgo) return dl;
  }
  if (rightgo) {
    dr = query(2 * v + 1, m + 1, r, max(ql, m + 1), qr);
    if (!leftgo) return dr;
  }
  D res = def;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++)
      for (int k = i; k <= j; k++)
        if (dl.m[i][k] + dr.m[k][j] < res.m[i][j])
          res.m[i][j] = dl.m[i][k] + dr.m[k][j];
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) def.m[i][j] = INF;
  fill(A, A + 2 * N, def);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    const int idx = i + N;
    for (int j = 0; j < 5; j++) A[idx].m[j][j] = 0;
    char c;
    cin >> c;
    switch (c) {
      case '2':
        A[idx].m[0][0] = 1;
        A[idx].m[0][1] = 0;
        break;
      case '0':
        A[idx].m[1][1] = 1;
        A[idx].m[1][2] = 0;
        break;
      case '1':
        A[idx].m[2][2] = 1;
        A[idx].m[2][3] = 0;
        break;
      case '7':
        A[idx].m[3][3] = 1;
        A[idx].m[3][4] = 0;
        break;
      case '6':
        A[idx].m[3][3] = 1;
        A[idx].m[4][4] = 1;
        break;
    }
  }
  for (int idx = N + n; idx < 2 * N; idx++)
    for (int i = 0; i < 5; i++) A[idx].m[i][i] = 0;
  for (int idx = N - 1; idx >= 1; idx--)
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++)
        for (int k = i; k <= j; k++)
          if (A[2 * idx].m[i][k] + A[2 * idx + 1].m[k][j] < A[idx].m[i][j])
            A[idx].m[i][j] = A[2 * idx].m[i][k] + A[2 * idx + 1].m[k][j];
  for (int i = 0; i < q; i++) {
    int a, b;
    cin >> a >> b;
    const int res = query(1, 0, N - 1, a - 1, b - 1).m[0][4];
    cout << (res == INF ? -1 : res) << '\n';
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 18;
const int INF = 2 * N;
struct D {
  int m[5][5];
};
D A[N * 2];
D def;
D query(int v, int l, int r, int ql, int qr) {
  if (l == ql && r == qr) return A[v];
  const int m = (l + r) / 2;
  const bool leftgo = ql <= m;
  const bool rightgo = qr > m;
  D dl, dr;
  if (leftgo) {
    dl = query(2 * v, l, m, ql, min(qr, m));
    if (!rightgo) return dl;
  }
  if (rightgo) {
    dr = query(2 * v + 1, m + 1, r, max(ql, m + 1), qr);
    if (!leftgo) return dr;
  }
  D res = def;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++)
      for (int k = i; k <= j; k++)
        if (dl.m[i][k] + dr.m[k][j] < res.m[i][j])
          res.m[i][j] = dl.m[i][k] + dr.m[k][j];
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) def.m[i][j] = INF;
  fill(A, A + 2 * N, def);
  int n, q;
  cin >> n >> q;
  for (int i = 0; i < n; i++) {
    const int idx = i + N;
    for (int j = 0; j < 5; j++) A[idx].m[j][j] = 0;
    char c;
    cin >> c;
    switch (c) {
      case '2':
        A[idx].m[0][0] = 1;
        A[idx].m[0][1] = 0;
        break;
      case '0':
        A[idx].m[1][1] = 1;
        A[idx].m[1][2] = 0;
        break;
      case '1':
        A[idx].m[2][2] = 1;
        A[idx].m[2][3] = 0;
        break;
      case '7':
        A[idx].m[3][3] = 1;
        A[idx].m[3][4] = 0;
        break;
      case '6':
        A[idx].m[3][3] = 1;
        A[idx].m[4][4] = 1;
        break;
    }
  }
  for (int idx = N + n; idx < 2 * N; idx++)
    for (int i = 0; i < 5; i++) A[idx].m[i][i] = 0;
  for (int idx = N - 1; idx >= 1; idx--)
    for (int i = 0; i < 5; i++)
      for (int j = i; j < 5; j++)
        for (int k = i; k <= j; k++)
          if (A[2 * idx].m[i][k] + A[2 * idx + 1].m[k][j] < A[idx].m[i][j])
            A[idx].m[i][j] = A[2 * idx].m[i][k] + A[2 * idx + 1].m[k][j];
  for (int i = 0; i < q; i++) {
    int a, b;
    cin >> a >> b;
    const int res = query(1, 0, N - 1, a - 1, b - 1).m[0][4];
    cout << (res == INF ? -1 : res) << '\n';
  }
  return 0;
}

```