#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int N = 200005;
inline int read() {
  int x = 0, f = 1;
  char c = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') x = 10 * x + c - '0', c = getchar();
  return x * f;
}
struct Matrix {
  int m[5][5];
  inline int *operator[](int id) { return m[id]; }
  inline Matrix operator*(Matrix rsh) {
    Matrix ret;
    memset(ret.m, 0x3f, sizeof(ret.m));
    for (register int i = 0; i < 5; i++)
      for (register int k = 0; k < 5; k++)
        for (register int j = 0; j < 5; j++)
          ret[i][j] = min(ret[i][j], m[i][k] + rsh[k][j]);
    return ret;
  }
} tr[N << 1], mdl[6];
inline void Pushup(int l, int mid, int r) {
  tr[(((l) + (r)) | ((l) != (r)))] =
      tr[(((mid + 1) + (r)) | ((mid + 1) != (r)))] *
      tr[(((l) + (mid)) | ((l) != (mid)))];
}
void Build(int l, int r) {
  if (l == r) {
    switch (getchar()) {
      case '2':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[1];
        break;
      case '0':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[2];
        break;
      case '1':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[3];
        break;
      case '7':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[4];
        break;
      case '6':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[5];
        break;
      default:
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[0];
        break;
    }
    return;
  }
  int mid = (l + r) >> 1;
  Build(l, mid);
  Build(mid + 1, r);
  Pushup(l, mid, r);
}
Matrix Query(int l, int r, int L, int R) {
  if (L <= l && r <= R) return tr[(((l) + (r)) | ((l) != (r)))];
  int mid = (l + r) >> 1;
  if (R <= mid) return Query(l, mid, L, R);
  if (L > mid) return Query(mid + 1, r, L, R);
  return Query(mid + 1, r, L, R) * Query(l, mid, L, R);
}
void Init() {
  memset(mdl[0].m, 0x3f, sizeof(mdl[0].m));
  for (int i = 0; i < 5; i++) mdl[0][i][i] = 0;
  for (int i = 1; i <= 5; i++) mdl[i] = mdl[0];
  mdl[1][0][0] = 1, mdl[1][1][0] = 0;
  mdl[2][1][1] = 1, mdl[2][2][1] = 0;
  mdl[3][2][2] = 1, mdl[3][3][2] = 0;
  mdl[4][3][3] = 1, mdl[4][4][3] = 0;
  mdl[5][4][4] = 1, mdl[5][3][3] = 1;
}
int n, q;
int x, y;
int main() {
  Init();
  n = read();
  q = read();
  Build(1, n);
  Matrix tmp;
  while (q--) {
    x = read(), y = read();
    tmp = Query(1, n, x, y);
    if (tmp[4][0] > (y - x + 1))
      puts("-1");
    else
      printf("%d\n", tmp[4][0]);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
const int INF = 0x3f3f3f3f;
const int N = 200005;
inline int read() {
  int x = 0, f = 1;
  char c = 0;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while ('0' <= c && c <= '9') x = 10 * x + c - '0', c = getchar();
  return x * f;
}
struct Matrix {
  int m[5][5];
  inline int *operator[](int id) { return m[id]; }
  inline Matrix operator*(Matrix rsh) {
    Matrix ret;
    memset(ret.m, 0x3f, sizeof(ret.m));
    for (register int i = 0; i < 5; i++)
      for (register int k = 0; k < 5; k++)
        for (register int j = 0; j < 5; j++)
          ret[i][j] = min(ret[i][j], m[i][k] + rsh[k][j]);
    return ret;
  }
} tr[N << 1], mdl[6];
inline void Pushup(int l, int mid, int r) {
  tr[(((l) + (r)) | ((l) != (r)))] =
      tr[(((mid + 1) + (r)) | ((mid + 1) != (r)))] *
      tr[(((l) + (mid)) | ((l) != (mid)))];
}
void Build(int l, int r) {
  if (l == r) {
    switch (getchar()) {
      case '2':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[1];
        break;
      case '0':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[2];
        break;
      case '1':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[3];
        break;
      case '7':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[4];
        break;
      case '6':
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[5];
        break;
      default:
        tr[(((l) + (r)) | ((l) != (r)))] = mdl[0];
        break;
    }
    return;
  }
  int mid = (l + r) >> 1;
  Build(l, mid);
  Build(mid + 1, r);
  Pushup(l, mid, r);
}
Matrix Query(int l, int r, int L, int R) {
  if (L <= l && r <= R) return tr[(((l) + (r)) | ((l) != (r)))];
  int mid = (l + r) >> 1;
  if (R <= mid) return Query(l, mid, L, R);
  if (L > mid) return Query(mid + 1, r, L, R);
  return Query(mid + 1, r, L, R) * Query(l, mid, L, R);
}
void Init() {
  memset(mdl[0].m, 0x3f, sizeof(mdl[0].m));
  for (int i = 0; i < 5; i++) mdl[0][i][i] = 0;
  for (int i = 1; i <= 5; i++) mdl[i] = mdl[0];
  mdl[1][0][0] = 1, mdl[1][1][0] = 0;
  mdl[2][1][1] = 1, mdl[2][2][1] = 0;
  mdl[3][2][2] = 1, mdl[3][3][2] = 0;
  mdl[4][3][3] = 1, mdl[4][4][3] = 0;
  mdl[5][4][4] = 1, mdl[5][3][3] = 1;
}
int n, q;
int x, y;
int main() {
  Init();
  n = read();
  q = read();
  Build(1, n);
  Matrix tmp;
  while (q--) {
    x = read(), y = read();
    tmp = Query(1, n, x, y);
    if (tmp[4][0] > (y - x + 1))
      puts("-1");
    else
      printf("%d\n", tmp[4][0]);
  }
  return 0;
}

```