#include <bits/stdc++.h>
using namespace std;
struct data {
  int a[6][6];
} t[800500];
int n, m;
char s[205000];
data operator+(data a, data b) {
  data c;
  for (int i = 0; i <= 5; i++)
    for (int j = 0; j <= 5; j++) c.a[i][j] = 1e9;
  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.a[i][k] + b.a[k][j]);
  return c;
}
void build(int i, int l, int r) {
  int mid = (l + r) / 2;
  if (l == r) {
    for (int j = 0; j <= 5; j++)
      for (int k = 0; k <= 5; k++) t[i].a[j][k] = 1e9;
    for (int j = 0; j <= 5; j++) t[i].a[j][j] = 0;
    if (s[l] == '2') t[i].a[0][0] = 1, t[i].a[0][1] = 0;
    if (s[l] == '0') t[i].a[1][1] = 1, t[i].a[1][2] = 0;
    if (s[l] == '1') t[i].a[2][2] = 1, t[i].a[2][3] = 0;
    if (s[l] == '6')
      t[i].a[3][3] = t[i].a[5][5] = 1, t[i].a[3][4] = t[i].a[5][4] = 0;
    if (s[l] == '7') t[i].a[3][3] = 1, t[i].a[3][5] = 0;
    return;
  }
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  t[i] = t[i * 2] + t[i * 2 + 1];
}
data ask(int i, int l, int r, int L, int R) {
  if (L <= l && r <= R) return t[i];
  int mid = (l + r) / 2;
  if (R <= mid)
    return ask(i * 2, l, mid, L, R);
  else if (L > mid)
    return ask(i * 2 + 1, mid + 1, r, L, R);
  else
    return ask(i * 2, l, mid, L, R) + ask(i * 2 + 1, mid + 1, r, L, R);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = ask(1, 1, n, l, r).a[0][5];
    printf("%d\n", ans == 1e9 ? -1 : ans);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
struct data {
  int a[6][6];
} t[800500];
int n, m;
char s[205000];
data operator+(data a, data b) {
  data c;
  for (int i = 0; i <= 5; i++)
    for (int j = 0; j <= 5; j++) c.a[i][j] = 1e9;
  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.a[i][k] + b.a[k][j]);
  return c;
}
void build(int i, int l, int r) {
  int mid = (l + r) / 2;
  if (l == r) {
    for (int j = 0; j <= 5; j++)
      for (int k = 0; k <= 5; k++) t[i].a[j][k] = 1e9;
    for (int j = 0; j <= 5; j++) t[i].a[j][j] = 0;
    if (s[l] == '2') t[i].a[0][0] = 1, t[i].a[0][1] = 0;
    if (s[l] == '0') t[i].a[1][1] = 1, t[i].a[1][2] = 0;
    if (s[l] == '1') t[i].a[2][2] = 1, t[i].a[2][3] = 0;
    if (s[l] == '6')
      t[i].a[3][3] = t[i].a[5][5] = 1, t[i].a[3][4] = t[i].a[5][4] = 0;
    if (s[l] == '7') t[i].a[3][3] = 1, t[i].a[3][5] = 0;
    return;
  }
  build(i * 2, l, mid);
  build(i * 2 + 1, mid + 1, r);
  t[i] = t[i * 2] + t[i * 2 + 1];
}
data ask(int i, int l, int r, int L, int R) {
  if (L <= l && r <= R) return t[i];
  int mid = (l + r) / 2;
  if (R <= mid)
    return ask(i * 2, l, mid, L, R);
  else if (L > mid)
    return ask(i * 2 + 1, mid + 1, r, L, R);
  else
    return ask(i * 2, l, mid, L, R) + ask(i * 2 + 1, mid + 1, r, L, R);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  build(1, 1, n);
  for (int i = 1; i <= m; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    int ans = ask(1, 1, n, l, r).a[0][5];
    printf("%d\n", ans == 1e9 ? -1 : ans);
  }
  return 0;
}

```