#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 1e3;
struct Node {
  int a[5][5];
  int* operator[](int x) { return a[x]; }
  Node operator+(Node b) {
    Node 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[i][j] = min(c[i][j], a[i][k] + b[k][j]);
    return c;
  }
} tr[MAXN << 2];
char str[MAXN];
int sl, sr;
Node build(int id, int l, int r) {
  if (l == r) {
    Node& e = tr[id];
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) e[i][j] = (i == j) ? 0 : 0x3f3f3f3f;
    switch (str[l]) {
      case '2':
        e[0][0] = 1;
        e[0][1] = 0;
        break;
      case '0':
        e[1][1] = 1;
        e[1][2] = 0;
        break;
      case '1':
        e[2][2] = 1;
        e[2][3] = 0;
        break;
      case '7':
        e[3][3] = 1;
        e[3][4] = 0;
        break;
      case '6':
        e[3][3] = 1;
        e[4][4] = 1;
        break;
    }
  } else {
    int m = l + r >> 1;
    tr[id] = build(id << 1, l, m) + build(id << 1 | 1, m + 1, r);
  }
  return tr[id];
}
Node query(int id, int l, int r) {
  if (sl <= l && r <= sr) return tr[id];
  int m = l + r >> 1;
  if (m >= sr) return query(id << 1, l, m);
  if (m < sl) return query(id << 1 | 1, m + 1, r);
  return query(id << 1, l, m) + query(id << 1 | 1, m + 1, r);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", str + 1);
  build(1, 1, n);
  while (q--) {
    scanf("%d%d", &sl, &sr);
    int ans = query(1, 1, n)[0][4];
    if (ans == 0x3f3f3f3f)
      puts("-1");
    else
      printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 MAXN = 2e5 + 1e3;
struct Node {
  int a[5][5];
  int* operator[](int x) { return a[x]; }
  Node operator+(Node b) {
    Node 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[i][j] = min(c[i][j], a[i][k] + b[k][j]);
    return c;
  }
} tr[MAXN << 2];
char str[MAXN];
int sl, sr;
Node build(int id, int l, int r) {
  if (l == r) {
    Node& e = tr[id];
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) e[i][j] = (i == j) ? 0 : 0x3f3f3f3f;
    switch (str[l]) {
      case '2':
        e[0][0] = 1;
        e[0][1] = 0;
        break;
      case '0':
        e[1][1] = 1;
        e[1][2] = 0;
        break;
      case '1':
        e[2][2] = 1;
        e[2][3] = 0;
        break;
      case '7':
        e[3][3] = 1;
        e[3][4] = 0;
        break;
      case '6':
        e[3][3] = 1;
        e[4][4] = 1;
        break;
    }
  } else {
    int m = l + r >> 1;
    tr[id] = build(id << 1, l, m) + build(id << 1 | 1, m + 1, r);
  }
  return tr[id];
}
Node query(int id, int l, int r) {
  if (sl <= l && r <= sr) return tr[id];
  int m = l + r >> 1;
  if (m >= sr) return query(id << 1, l, m);
  if (m < sl) return query(id << 1 | 1, m + 1, r);
  return query(id << 1, l, m) + query(id << 1 | 1, m + 1, r);
}
int main() {
  int n, q;
  scanf("%d%d", &n, &q);
  scanf("%s", str + 1);
  build(1, 1, n);
  while (q--) {
    scanf("%d%d", &sl, &sr);
    int ans = query(1, 1, n)[0][4];
    if (ans == 0x3f3f3f3f)
      puts("-1");
    else
      printf("%d\n", ans);
  }
  return 0;
}

```