#include <bits/stdc++.h>
using namespace std;
int n, m;
char s1[200005];
struct Mac {
  int num[5][5];
  Mac() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) num[i][j] = 1e9;
  }
  Mac operator+(const Mac &c) const {
    Mac tmp;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        for (int k = 0; k < 5; k++)
          tmp.num[i][j] = min(tmp.num[i][j], num[i][k] + c.num[k][j]);
      }
    return tmp;
  }
} tree[200005 * 4];
void Build(int L, int R, int p) {
  if (L == R) {
    for (int i = 0; i < 5; i++) tree[p].num[i][i] = 0;
    if (s1[L] == '2') tree[p].num[0][0] = 1, tree[p].num[0][1] = 0;
    if (s1[L] == '0') tree[p].num[1][1] = 1, tree[p].num[1][2] = 0;
    if (s1[L] == '1') tree[p].num[2][2] = 1, tree[p].num[2][3] = 0;
    if (s1[L] == '7') tree[p].num[3][3] = 1, tree[p].num[3][4] = 0;
    if (s1[L] == '6') tree[p].num[3][3] = 1, tree[p].num[4][4] = 1;
    return;
  }
  int mid = (L + R) >> 1;
  Build(L, mid, p << 1);
  Build(mid + 1, R, p << 1 | 1);
  tree[p] = tree[p << 1] + tree[p << 1 | 1];
}
Mac Query(int L, int R, int l, int r, int p) {
  if (L == l && R == r) return tree[p];
  int mid = (L + R) >> 1;
  if (mid >= r)
    return Query(L, mid, l, r, p << 1);
  else if (mid < l)
    return Query(mid + 1, R, l, r, p << 1 | 1);
  else
    return Query(L, mid, l, mid, p << 1) +
           Query(mid + 1, R, mid + 1, r, p << 1 | 1);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s1 + 1);
  Build(1, n, 1);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    Mac res;
    res = Query(1, n, l, r, 1);
    if (res.num[0][4] == 1e9)
      puts("-1");
    else
      printf("%d\n", res.num[0][4]);
  }
  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;
int n, m;
char s1[200005];
struct Mac {
  int num[5][5];
  Mac() {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) num[i][j] = 1e9;
  }
  Mac operator+(const Mac &c) const {
    Mac tmp;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        for (int k = 0; k < 5; k++)
          tmp.num[i][j] = min(tmp.num[i][j], num[i][k] + c.num[k][j]);
      }
    return tmp;
  }
} tree[200005 * 4];
void Build(int L, int R, int p) {
  if (L == R) {
    for (int i = 0; i < 5; i++) tree[p].num[i][i] = 0;
    if (s1[L] == '2') tree[p].num[0][0] = 1, tree[p].num[0][1] = 0;
    if (s1[L] == '0') tree[p].num[1][1] = 1, tree[p].num[1][2] = 0;
    if (s1[L] == '1') tree[p].num[2][2] = 1, tree[p].num[2][3] = 0;
    if (s1[L] == '7') tree[p].num[3][3] = 1, tree[p].num[3][4] = 0;
    if (s1[L] == '6') tree[p].num[3][3] = 1, tree[p].num[4][4] = 1;
    return;
  }
  int mid = (L + R) >> 1;
  Build(L, mid, p << 1);
  Build(mid + 1, R, p << 1 | 1);
  tree[p] = tree[p << 1] + tree[p << 1 | 1];
}
Mac Query(int L, int R, int l, int r, int p) {
  if (L == l && R == r) return tree[p];
  int mid = (L + R) >> 1;
  if (mid >= r)
    return Query(L, mid, l, r, p << 1);
  else if (mid < l)
    return Query(mid + 1, R, l, r, p << 1 | 1);
  else
    return Query(L, mid, l, mid, p << 1) +
           Query(mid + 1, R, mid + 1, r, p << 1 | 1);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s1 + 1);
  Build(1, n, 1);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    Mac res;
    res = Query(1, n, l, r, 1);
    if (res.num[0][4] == 1e9)
      puts("-1");
    else
      printf("%d\n", res.num[0][4]);
  }
  return 0;
}

```