#include <bits/stdc++.h>
using namespace std;
int get(int n, int p) {
  int res = 0;
  while (n >= p) res += n / p, n /= p;
  return res;
}
template <class T>
T power(T N, T P) {
  return (P == 0) ? 1 : N * power(N, P - 1);
}
void VC(vector<long long int>& v) {
  for (int i = 0; i < v.size(); i++) printf("%lld ", v[i]);
  printf("\n");
}
template <typename T>
void print(const T& v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1& first, const T2&... rest) {
  print(first);
  print(rest...);
}
long long BigMod(long long int a, long long int b, int M) {
  long long remainder, answer = 1;
  remainder = a % M;
  while (b != 0) {
    if (b % 2 == 1) answer = (answer * remainder) % M;
    remainder = (remainder * remainder) % M;
    b /= 2;
  }
  return answer;
}
bool cmp(int a, int b) { return a > b; }
int cnt[102][1005] = {0};
int main() {
  int n, dd;
  scanf("%d", &(n));
  string s;
  cin >> s;
  int a[n + 4], start[n + 4];
  for (int i = 0; i < n; i++) {
    scanf("%d", &(a[i]));
    scanf("%d", &(start[i]));
  }
  for (int i = 0; i < n; i++) {
    cnt[i][0] = ((s[i] == '1') ? 1 : 0);
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      for (int j = 0; j < start[i]; j++) {
        cnt[i][j] = 1;
      }
      start[i] += a[i];
    } else {
      for (int j = 0; j < start[i]; j++) {
        cnt[i][j] = 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    dd = 1;
    for (int j = start[i]; j <= 1e3; j++, dd++) {
      cnt[i][j] = 1;
      if (dd % a[i] == 0) {
        j += a[i];
      }
    }
  }
  long long int mxx = 0, sum;
  for (int i = 0; i < 1e3; i++) {
    sum = 0;
    for (int j = 0; j < n; j++) {
      sum += cnt[j][i];
    }
    mxx = max(mxx, sum);
  }
  cout << mxx;
  return 0;
}

### Prompt

Create a solution in CPP for the following problem:

It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.

After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.

You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.

<image> Here is a graphic for the first example.

Input

The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.

The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".

The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.

Output

Print a single integer — the maximum number of lights that will ever be on at the same time.

Examples

Input


3
101
3 3
3 2
3 1


Output


2


Input


4
1111
3 4
5 2
3 1
3 2


Output


4


Input


6
011100
5 3
5 5
2 4
3 5
4 2
1 5


Output


6

Note

For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).

In the second example, all lights are initially on. So the answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int get(int n, int p) {
  int res = 0;
  while (n >= p) res += n / p, n /= p;
  return res;
}
template <class T>
T power(T N, T P) {
  return (P == 0) ? 1 : N * power(N, P - 1);
}
void VC(vector<long long int>& v) {
  for (int i = 0; i < v.size(); i++) printf("%lld ", v[i]);
  printf("\n");
}
template <typename T>
void print(const T& v) {
  cerr << v << ' ';
}
template <typename T1, typename... T2>
void print(const T1& first, const T2&... rest) {
  print(first);
  print(rest...);
}
long long BigMod(long long int a, long long int b, int M) {
  long long remainder, answer = 1;
  remainder = a % M;
  while (b != 0) {
    if (b % 2 == 1) answer = (answer * remainder) % M;
    remainder = (remainder * remainder) % M;
    b /= 2;
  }
  return answer;
}
bool cmp(int a, int b) { return a > b; }
int cnt[102][1005] = {0};
int main() {
  int n, dd;
  scanf("%d", &(n));
  string s;
  cin >> s;
  int a[n + 4], start[n + 4];
  for (int i = 0; i < n; i++) {
    scanf("%d", &(a[i]));
    scanf("%d", &(start[i]));
  }
  for (int i = 0; i < n; i++) {
    cnt[i][0] = ((s[i] == '1') ? 1 : 0);
  }
  for (int i = 0; i < n; i++) {
    if (s[i] == '1') {
      for (int j = 0; j < start[i]; j++) {
        cnt[i][j] = 1;
      }
      start[i] += a[i];
    } else {
      for (int j = 0; j < start[i]; j++) {
        cnt[i][j] = 0;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    dd = 1;
    for (int j = start[i]; j <= 1e3; j++, dd++) {
      cnt[i][j] = 1;
      if (dd % a[i] == 0) {
        j += a[i];
      }
    }
  }
  long long int mxx = 0, sum;
  for (int i = 0; i < 1e3; i++) {
    sum = 0;
    for (int j = 0; j < n; j++) {
      sum += cnt[j][i];
    }
    mxx = max(mxx, sum);
  }
  cout << mxx;
  return 0;
}

```