#include <bits/stdc++.h>
using namespace std;
typedef int State[5][5];
const int N = 1 << 18;
const int INF = 1e8;
State tr[2 * N];
char s[N];
int n;
void resetState(State a) {
  fill_n((int*)a, 25, INF);
  for (int i = 0; i < 5; i++) {
    a[i][i] = 0;
  }
}
void setState(State a, char ch) {
  resetState(a);
  switch (ch) {
    case '0':
      a[1][1] = 1;
      a[1][2] = 0;
      break;
    case '1':
      a[2][2] = 1;
      a[2][3] = 0;
      break;
    case '2':
      a[0][0] = 1;
      a[0][1] = 0;
      break;
    case '6':
      a[3][3] = 1;
      a[4][4] = 1;
      break;
    case '7':
      a[3][3] = 1;
      a[3][4] = 0;
      break;
  }
}
void merge(State res, State left, State right) {
  int i, j, k;
  State tmp;
  for (i = 0; i < 5; i++) {
    for (j = i; j < 5; j++) {
      tmp[i][j] = INF;
      for (k = i; k <= j; k++) {
        tmp[i][j] = min(tmp[i][j], left[i][k] + right[k][j]);
      }
    }
  }
  memcpy(res, tmp, sizeof(State));
}
void buildSegTree() {
  int i;
  for (i = 0; i < N; i++) {
    if (i < n) {
      setState(tr[i + N], s[i]);
    } else {
      resetState(tr[i + N]);
    }
  }
  for (i = N - 1; i >= 1; i--) {
    merge(tr[i], tr[2 * i], tr[2 * i + 1]);
  }
}
int query(int left, int right) {
  State lState, rState;
  int l, r;
  resetState(lState);
  resetState(rState);
  for (l = left + N, r = right + N + 1; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      merge(lState, lState, tr[l++]);
    }
    if (r & 1) {
      merge(rState, tr[--r], rState);
    }
  }
  State res;
  merge(res, lState, rState);
  return res[0][4] >= INF ? -1 : res[0][4];
}
int l[N];
int r[N];
int ans[N];
int main() {
  int i, qNum;
  tr[10][0][0] = 10;
  scanf("%d%d", &n, &qNum);
  scanf("%s", s);
  for (i = 1; i <= qNum; i++) {
    scanf("%d%d", &l[i], &r[i]);
    l[i]--;
    r[i]--;
  }
  buildSegTree();
  for (i = 1; i <= qNum; i++) {
    ans[i] = query(l[i], r[i]);
  }
  for (i = 1; i <= qNum; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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;
typedef int State[5][5];
const int N = 1 << 18;
const int INF = 1e8;
State tr[2 * N];
char s[N];
int n;
void resetState(State a) {
  fill_n((int*)a, 25, INF);
  for (int i = 0; i < 5; i++) {
    a[i][i] = 0;
  }
}
void setState(State a, char ch) {
  resetState(a);
  switch (ch) {
    case '0':
      a[1][1] = 1;
      a[1][2] = 0;
      break;
    case '1':
      a[2][2] = 1;
      a[2][3] = 0;
      break;
    case '2':
      a[0][0] = 1;
      a[0][1] = 0;
      break;
    case '6':
      a[3][3] = 1;
      a[4][4] = 1;
      break;
    case '7':
      a[3][3] = 1;
      a[3][4] = 0;
      break;
  }
}
void merge(State res, State left, State right) {
  int i, j, k;
  State tmp;
  for (i = 0; i < 5; i++) {
    for (j = i; j < 5; j++) {
      tmp[i][j] = INF;
      for (k = i; k <= j; k++) {
        tmp[i][j] = min(tmp[i][j], left[i][k] + right[k][j]);
      }
    }
  }
  memcpy(res, tmp, sizeof(State));
}
void buildSegTree() {
  int i;
  for (i = 0; i < N; i++) {
    if (i < n) {
      setState(tr[i + N], s[i]);
    } else {
      resetState(tr[i + N]);
    }
  }
  for (i = N - 1; i >= 1; i--) {
    merge(tr[i], tr[2 * i], tr[2 * i + 1]);
  }
}
int query(int left, int right) {
  State lState, rState;
  int l, r;
  resetState(lState);
  resetState(rState);
  for (l = left + N, r = right + N + 1; l < r; l >>= 1, r >>= 1) {
    if (l & 1) {
      merge(lState, lState, tr[l++]);
    }
    if (r & 1) {
      merge(rState, tr[--r], rState);
    }
  }
  State res;
  merge(res, lState, rState);
  return res[0][4] >= INF ? -1 : res[0][4];
}
int l[N];
int r[N];
int ans[N];
int main() {
  int i, qNum;
  tr[10][0][0] = 10;
  scanf("%d%d", &n, &qNum);
  scanf("%s", s);
  for (i = 1; i <= qNum; i++) {
    scanf("%d%d", &l[i], &r[i]);
    l[i]--;
    r[i]--;
  }
  buildSegTree();
  for (i = 1; i <= qNum; i++) {
    ans[i] = query(l[i], r[i]);
  }
  for (i = 1; i <= qNum; i++) {
    printf("%d\n", ans[i]);
  }
  return 0;
}

```