#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
struct mat {
  int a[5][5];
  mat operator+(const mat &rhs) const {
    mat c;
    memset(c.a, 0x3f, sizeof(c.a));
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        for (int k = 0; k < 5; k++) {
          c.a[i][j] = min(c.a[i][j], a[i][k] + rhs.a[k][j]);
        }
      }
    }
    return c;
  }
};
const int maxn = 2e5 + 10;
mat dat[1 << 19];
char s[maxn];
int n, m;
struct tree {
  int qL, qR;
  void build(int o, int L, int R) {
    if (L == R) {
      for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++) dat[o].a[i][j] = (i == j) ? 0 : INF;
      switch (s[L]) {
        case '2':
          dat[o].a[0][1] = 0;
          dat[o].a[0][0] = 1;
          break;
        case '0':
          dat[o].a[1][2] = 0;
          dat[o].a[1][1] = 1;
          break;
        case '1':
          dat[o].a[2][2] = 1;
          dat[o].a[2][3] = 0;
          break;
        case '6':
          dat[o].a[3][3] = 1;
          dat[o].a[4][4] = 1;
          break;
        case '7':
          dat[o].a[3][4] = 0;
          dat[o].a[3][3] = 1;
          break;
      }
      return;
    }
    build((o << 1), L, (L + (R - L) / 2));
    build(((o << 1) | 1), (L + (R - L) / 2) + 1, R);
    dat[o] = dat[(o << 1)] + dat[((o << 1) | 1)];
  }
  mat query(int o, int L, int R) {
    if (qL <= L && R <= qR) return dat[o];
    if (qR <= (L + (R - L) / 2)) return query((o << 1), L, (L + (R - L) / 2));
    if (qL > (L + (R - L) / 2))
      return query(((o << 1) | 1), (L + (R - L) / 2) + 1, R);
    return query((o << 1), L, (L + (R - L) / 2)) +
           query(((o << 1) | 1), (L + (R - L) / 2) + 1, R);
  }
} t;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  t.build(1, 1, n);
  while (m--) {
    scanf("%d%d", &t.qL, &t.qR);
    int ans = t.query(1, 1, n).a[0][4];
    if (ans >= INF) ans = -1;
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Generate 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;
struct mat {
  int a[5][5];
  mat operator+(const mat &rhs) const {
    mat c;
    memset(c.a, 0x3f, sizeof(c.a));
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        for (int k = 0; k < 5; k++) {
          c.a[i][j] = min(c.a[i][j], a[i][k] + rhs.a[k][j]);
        }
      }
    }
    return c;
  }
};
const int maxn = 2e5 + 10;
mat dat[1 << 19];
char s[maxn];
int n, m;
struct tree {
  int qL, qR;
  void build(int o, int L, int R) {
    if (L == R) {
      for (int i = 0; i < 5; i++)
        for (int j = 0; j < 5; j++) dat[o].a[i][j] = (i == j) ? 0 : INF;
      switch (s[L]) {
        case '2':
          dat[o].a[0][1] = 0;
          dat[o].a[0][0] = 1;
          break;
        case '0':
          dat[o].a[1][2] = 0;
          dat[o].a[1][1] = 1;
          break;
        case '1':
          dat[o].a[2][2] = 1;
          dat[o].a[2][3] = 0;
          break;
        case '6':
          dat[o].a[3][3] = 1;
          dat[o].a[4][4] = 1;
          break;
        case '7':
          dat[o].a[3][4] = 0;
          dat[o].a[3][3] = 1;
          break;
      }
      return;
    }
    build((o << 1), L, (L + (R - L) / 2));
    build(((o << 1) | 1), (L + (R - L) / 2) + 1, R);
    dat[o] = dat[(o << 1)] + dat[((o << 1) | 1)];
  }
  mat query(int o, int L, int R) {
    if (qL <= L && R <= qR) return dat[o];
    if (qR <= (L + (R - L) / 2)) return query((o << 1), L, (L + (R - L) / 2));
    if (qL > (L + (R - L) / 2))
      return query(((o << 1) | 1), (L + (R - L) / 2) + 1, R);
    return query((o << 1), L, (L + (R - L) / 2)) +
           query(((o << 1) | 1), (L + (R - L) / 2) + 1, R);
  }
} t;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  t.build(1, 1, n);
  while (m--) {
    scanf("%d%d", &t.qL, &t.qR);
    int ans = t.query(1, 1, n).a[0][4];
    if (ans >= INF) ans = -1;
    printf("%d\n", ans);
  }
  return 0;
}

```