#include <bits/stdc++.h>
using namespace std;
const int kInf = 0x3f3f3f3f, kN = 200005;
int n, q;
string s;
struct Node {
  int a[5][5];
  Node operator+(const Node &b) const {
    Node c;
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) {
        c.a[i][j] = kInf;
        for (int k = 0; k < 5; ++k)
          c.a[i][j] = min(c.a[i][j], a[i][k] + b.a[k][j]);
      }
    return c;
  }
} tr[kN << 2];
void build(int x, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) tr[x].a[i][j] = (i == j) ? 0 : kInf;
    if (s[l] == '2')
      tr[x].a[0][0] = 1, tr[x].a[0][1] = 0;
    else if (s[l] == '0')
      tr[x].a[1][1] = 1, tr[x].a[1][2] = 0;
    else if (s[l] == '1')
      tr[x].a[2][2] = 1, tr[x].a[2][3] = 0;
    else if (s[l] == '7')
      tr[x].a[3][3] = 1, tr[x].a[3][4] = 0;
    else if (s[l] == '6')
      tr[x].a[3][3] = 1, tr[x].a[4][4] = 1;
    return;
  } else {
    int mid = (l + r) >> 1;
    build(x << 1, l, mid);
    build(x << 1 | 1, mid + 1, r);
    tr[x] = tr[x << 1] + tr[x << 1 | 1];
  }
}
Node query(int x, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return tr[x];
  int mid = (l + r) >> 1;
  if (qr <= mid) return query(x << 1, l, mid, ql, qr);
  if (ql > mid) return query(x << 1 | 1, mid + 1, r, ql, qr);
  return query(x << 1, l, mid, ql, qr) + query(x << 1 | 1, mid + 1, r, ql, qr);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q >> s;
  s = ' ' + s;
  build(1, 1, n);
  while (q--) {
    int l, r;
    cin >> l >> r;
    int ans = query(1, 1, n, l, r).a[0][4];
    if (ans == kInf)
      cout << -1 << "\n";
    else
      cout << ans << "\n";
  }
  cout << flush;
  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 kInf = 0x3f3f3f3f, kN = 200005;
int n, q;
string s;
struct Node {
  int a[5][5];
  Node operator+(const Node &b) const {
    Node c;
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) {
        c.a[i][j] = kInf;
        for (int k = 0; k < 5; ++k)
          c.a[i][j] = min(c.a[i][j], a[i][k] + b.a[k][j]);
      }
    return c;
  }
} tr[kN << 2];
void build(int x, int l, int r) {
  if (l == r) {
    for (int i = 0; i < 5; ++i)
      for (int j = 0; j < 5; ++j) tr[x].a[i][j] = (i == j) ? 0 : kInf;
    if (s[l] == '2')
      tr[x].a[0][0] = 1, tr[x].a[0][1] = 0;
    else if (s[l] == '0')
      tr[x].a[1][1] = 1, tr[x].a[1][2] = 0;
    else if (s[l] == '1')
      tr[x].a[2][2] = 1, tr[x].a[2][3] = 0;
    else if (s[l] == '7')
      tr[x].a[3][3] = 1, tr[x].a[3][4] = 0;
    else if (s[l] == '6')
      tr[x].a[3][3] = 1, tr[x].a[4][4] = 1;
    return;
  } else {
    int mid = (l + r) >> 1;
    build(x << 1, l, mid);
    build(x << 1 | 1, mid + 1, r);
    tr[x] = tr[x << 1] + tr[x << 1 | 1];
  }
}
Node query(int x, int l, int r, int ql, int qr) {
  if (ql <= l && r <= qr) return tr[x];
  int mid = (l + r) >> 1;
  if (qr <= mid) return query(x << 1, l, mid, ql, qr);
  if (ql > mid) return query(x << 1 | 1, mid + 1, r, ql, qr);
  return query(x << 1, l, mid, ql, qr) + query(x << 1 | 1, mid + 1, r, ql, qr);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> q >> s;
  s = ' ' + s;
  build(1, 1, n);
  while (q--) {
    int l, r;
    cin >> l >> r;
    int ans = query(1, 1, n, l, r).a[0][4];
    if (ans == kInf)
      cout << -1 << "\n";
    else
      cout << ans << "\n";
  }
  cout << flush;
  return 0;
}

```