#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T _abs(T a) {
  return a > 0 ? a : -a;
}
const int maxn = 200010;
const int inf = (1 << 20);
int n, m;
char c[maxn];
struct node {
  int l, r, mid, a[5][5];
} t[maxn * 4];
void build(int l, int r, int k) {
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t[k].a[i][j] = inf;
  t[k].l = l;
  t[k].r = r;
  t[k].mid = (l + r) / 2;
  if (l == r) {
    for (int j = 0; j < 5; j++) t[k].a[j][j] = 0;
    if (c[l] == '2') {
      t[k].a[0][0] = 1;
      t[k].a[0][1] = 0;
    } else if (c[l] == '0') {
      t[k].a[1][1] = 1;
      t[k].a[1][2] = 0;
    } else if (c[l] == '1') {
      t[k].a[2][2] = 1;
      t[k].a[2][3] = 0;
    } else if (c[l] == '7') {
      t[k].a[3][3] = 1;
      t[k].a[3][4] = 0;
    } else if (c[l] == '6') {
      t[k].a[3][3] = 1;
      t[k].a[4][4] = 1;
    }
  } else {
    int mid = (l + r) / 2;
    build(l, mid, 2 * k);
    build(mid + 1, r, 2 * k + 1);
    node n;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        n.a[i][j] = inf;
        for (int p = 0; p < 5; p++)
          n.a[i][j] = min(n.a[i][j], t[2 * k].a[i][p] + t[2 * k + 1].a[p][j]);
        t[k].a[i][j] = n.a[i][j];
      }
  }
}
node query(int l, int r, int k) {
  if (t[k].l == l && t[k].r == r)
    return t[k];
  else if (r <= t[k].mid)
    return query(l, r, 2 * k);
  else if (l > t[k].mid)
    return query(l, r, 2 * k + 1);
  else {
    node n1 = query(l, t[k].mid, 2 * k), n2 = query(t[k].mid + 1, r, 2 * k + 1);
    node n;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        n.a[i][j] = inf;
        for (int p = 0; p < 5; p++)
          n.a[i][j] = min(n.a[i][j], n1.a[i][p] + n2.a[p][j]);
      }
    return n;
  }
}
int main() {
  int x, y;
  while (~scanf("%d%d", &n, &m)) {
    scanf("%s", c + 1);
    build(1, n, 1);
    while (m--) {
      scanf("%d%d", &x, &y);
      node n = query(x, y, 1);
      printf("%d\n", n.a[0][4] >= inf ? -1 : n.a[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;
#pragma comment(linker, "/STACK:102400000,102400000")
template <typename T>
T gcd(T a, T b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a / gcd(a, b) * b;
}
template <typename T>
T _abs(T a) {
  return a > 0 ? a : -a;
}
const int maxn = 200010;
const int inf = (1 << 20);
int n, m;
char c[maxn];
struct node {
  int l, r, mid, a[5][5];
} t[maxn * 4];
void build(int l, int r, int k) {
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t[k].a[i][j] = inf;
  t[k].l = l;
  t[k].r = r;
  t[k].mid = (l + r) / 2;
  if (l == r) {
    for (int j = 0; j < 5; j++) t[k].a[j][j] = 0;
    if (c[l] == '2') {
      t[k].a[0][0] = 1;
      t[k].a[0][1] = 0;
    } else if (c[l] == '0') {
      t[k].a[1][1] = 1;
      t[k].a[1][2] = 0;
    } else if (c[l] == '1') {
      t[k].a[2][2] = 1;
      t[k].a[2][3] = 0;
    } else if (c[l] == '7') {
      t[k].a[3][3] = 1;
      t[k].a[3][4] = 0;
    } else if (c[l] == '6') {
      t[k].a[3][3] = 1;
      t[k].a[4][4] = 1;
    }
  } else {
    int mid = (l + r) / 2;
    build(l, mid, 2 * k);
    build(mid + 1, r, 2 * k + 1);
    node n;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        n.a[i][j] = inf;
        for (int p = 0; p < 5; p++)
          n.a[i][j] = min(n.a[i][j], t[2 * k].a[i][p] + t[2 * k + 1].a[p][j]);
        t[k].a[i][j] = n.a[i][j];
      }
  }
}
node query(int l, int r, int k) {
  if (t[k].l == l && t[k].r == r)
    return t[k];
  else if (r <= t[k].mid)
    return query(l, r, 2 * k);
  else if (l > t[k].mid)
    return query(l, r, 2 * k + 1);
  else {
    node n1 = query(l, t[k].mid, 2 * k), n2 = query(t[k].mid + 1, r, 2 * k + 1);
    node n;
    for (int i = 0; i < 5; i++)
      for (int j = 0; j < 5; j++) {
        n.a[i][j] = inf;
        for (int p = 0; p < 5; p++)
          n.a[i][j] = min(n.a[i][j], n1.a[i][p] + n2.a[p][j]);
      }
    return n;
  }
}
int main() {
  int x, y;
  while (~scanf("%d%d", &n, &m)) {
    scanf("%s", c + 1);
    build(1, n, 1);
    while (m--) {
      scanf("%d%d", &x, &y);
      node n = query(x, y, 1);
      printf("%d\n", n.a[0][4] >= inf ? -1 : n.a[0][4]);
    }
  }
  return 0;
}

```