#include <bits/stdc++.h>
using namespace std;
const int INF = 1e6;
const int N = 200010;
int n, q, f[N][10], sum[N][10], g[3][N], t2, t0, t1, t7, t, x, y, ans;
char s[N];
pair<int, int> p[N];
struct Point {
  int c[5][5];
};
struct Node {
  int l, r, c[5][5];
} node[N * 4];
void build(int x, int l, int r) {
  node[x].l = l, node[x].r = r;
  Node &t = node[x];
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t.c[i][j] = INF;
  if (l == r) {
    for (int i = 0; i < 5; i++) t.c[i][i] = 0;
    if (s[l] == '2')
      t.c[0][1] = 0, t.c[0][0] = 1;
    else if (s[l] == '0')
      t.c[1][2] = 0, t.c[1][1] = 1;
    else if (s[l] == '1')
      t.c[2][3] = 0, t.c[2][2] = 1;
    else if (s[l] == '7')
      t.c[3][4] = 0, t.c[3][3] = 1;
    else if (s[l] == '6')
      t.c[3][3] = 1, t.c[4][4] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(x * 2, l, mid);
  build(x * 2 + 1, mid + 1, r);
  Node &L = node[x * 2], &R = node[x * 2 + 1];
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++)
        t.c[i][j] = min(t.c[i][j], L.c[i][k] + R.c[k][j]);
    }
}
Point find(int x, int l, int r) {
  Point ret;
  Point &t = ret;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t.c[i][j] = INF;
  if (node[x].l == l && node[x].r == r) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) ret.c[i][j] = node[x].c[i][j];
    return ret;
  }
  int mid = (node[x].l + node[x].r) / 2;
  if (r <= mid)
    return find(x * 2, l, r);
  else if (l > mid)
    return find(x * 2 + 1, l, r);
  Point ret1 = find(x * 2, l, mid), ret2 = find(x * 2 + 1, mid + 1, r);
  Point &L = ret1, &R = ret2;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++)
        t.c[i][j] = min(t.c[i][j], L.c[i][k] + R.c[k][j]);
    }
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  build(1, 1, n);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &x, &y);
    Point p = find(1, x, y);
    if (p.c[0][4] >= INF)
      puts("-1");
    else
      printf("%d\n", p.c[0][4]);
  }
  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;
const int INF = 1e6;
const int N = 200010;
int n, q, f[N][10], sum[N][10], g[3][N], t2, t0, t1, t7, t, x, y, ans;
char s[N];
pair<int, int> p[N];
struct Point {
  int c[5][5];
};
struct Node {
  int l, r, c[5][5];
} node[N * 4];
void build(int x, int l, int r) {
  node[x].l = l, node[x].r = r;
  Node &t = node[x];
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t.c[i][j] = INF;
  if (l == r) {
    for (int i = 0; i < 5; i++) t.c[i][i] = 0;
    if (s[l] == '2')
      t.c[0][1] = 0, t.c[0][0] = 1;
    else if (s[l] == '0')
      t.c[1][2] = 0, t.c[1][1] = 1;
    else if (s[l] == '1')
      t.c[2][3] = 0, t.c[2][2] = 1;
    else if (s[l] == '7')
      t.c[3][4] = 0, t.c[3][3] = 1;
    else if (s[l] == '6')
      t.c[3][3] = 1, t.c[4][4] = 1;
    return;
  }
  int mid = (l + r) / 2;
  build(x * 2, l, mid);
  build(x * 2 + 1, mid + 1, r);
  Node &L = node[x * 2], &R = node[x * 2 + 1];
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++)
        t.c[i][j] = min(t.c[i][j], L.c[i][k] + R.c[k][j]);
    }
}
Point find(int x, int l, int r) {
  Point ret;
  Point &t = ret;
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t.c[i][j] = INF;
  if (node[x].l == l && node[x].r == r) {
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) ret.c[i][j] = node[x].c[i][j];
    return ret;
  }
  int mid = (node[x].l + node[x].r) / 2;
  if (r <= mid)
    return find(x * 2, l, r);
  else if (l > mid)
    return find(x * 2 + 1, l, r);
  Point ret1 = find(x * 2, l, mid), ret2 = find(x * 2 + 1, mid + 1, r);
  Point &L = ret1, &R = ret2;
  for (int i = 0; i < 5; i++)
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++)
        t.c[i][j] = min(t.c[i][j], L.c[i][k] + R.c[k][j]);
    }
  return ret;
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", s + 1);
  build(1, 1, n);
  for (int i = 0; i < q; i++) {
    scanf("%d%d", &x, &y);
    Point p = find(1, x, y);
    if (p.c[0][4] >= INF)
      puts("-1");
    else
      printf("%d\n", p.c[0][4]);
  }
  return 0;
}

```