#include <bits/stdc++.h>
int Gcd(int a[], int n);
int Lcm(int a[], int n);
int main() {
  int n, initial[100], a[100], b[100], a1[100], i, j, lcm, counter = 0, mod,
                                                           max = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%1d", &initial[i]);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    a[i] *= 2;
    a1[i] = a[i];
  }
  for (int t = 0; t < 1000; t++) {
    counter = 0;
    for (j = 0; j < n; j++) {
      mod = (t >= b[j]) ? ((t - b[j])) % a[j] : -1;
      if ((!initial[j] && (mod >= 0 && (mod < a[j] / 2))) ||
          (initial[j] && (mod == -1 || ((mod >= a[j] / 2) && mod < a[j]))))
        counter++;
    }
    max = (counter > max) ? counter : max;
  }
  printf("%d\n", max);
  return 0;
}
int Gcd(int a[], int n) {
  int min, pos = 0, counter = 0, flag[100] = {0};
  while (counter < n - 1) {
    for (int i = 0; i < n; i++)
      if (a[i] && a[i] < a[pos]) pos = i;
    min = a[pos];
    for (int i = 0; i < n; i++) {
      if (i == pos) continue;
      a[i] = a[i] % min;
      if (!a[i] && !flag[i]) {
        counter++;
        flag[i] = 1;
      }
    }
  }
  for (int i = 0; i < n; i++)
    if (a[i]) return a[i];
}
int Lcm(int a[], int n) {
  int m = 1;
  for (int i = 0; i < n; i++) m *= a[i];
  for (int i = 0; i < n; i++) a[i] = m / a[i];
  int gcd = Gcd(a, n);
  return m / gcd;
}

### 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>
int Gcd(int a[], int n);
int Lcm(int a[], int n);
int main() {
  int n, initial[100], a[100], b[100], a1[100], i, j, lcm, counter = 0, mod,
                                                           max = 0;
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%1d", &initial[i]);
  for (i = 0; i < n; i++) {
    scanf("%d %d", &a[i], &b[i]);
    a[i] *= 2;
    a1[i] = a[i];
  }
  for (int t = 0; t < 1000; t++) {
    counter = 0;
    for (j = 0; j < n; j++) {
      mod = (t >= b[j]) ? ((t - b[j])) % a[j] : -1;
      if ((!initial[j] && (mod >= 0 && (mod < a[j] / 2))) ||
          (initial[j] && (mod == -1 || ((mod >= a[j] / 2) && mod < a[j]))))
        counter++;
    }
    max = (counter > max) ? counter : max;
  }
  printf("%d\n", max);
  return 0;
}
int Gcd(int a[], int n) {
  int min, pos = 0, counter = 0, flag[100] = {0};
  while (counter < n - 1) {
    for (int i = 0; i < n; i++)
      if (a[i] && a[i] < a[pos]) pos = i;
    min = a[pos];
    for (int i = 0; i < n; i++) {
      if (i == pos) continue;
      a[i] = a[i] % min;
      if (!a[i] && !flag[i]) {
        counter++;
        flag[i] = 1;
      }
    }
  }
  for (int i = 0; i < n; i++)
    if (a[i]) return a[i];
}
int Lcm(int a[], int n) {
  int m = 1;
  for (int i = 0; i < n; i++) m *= a[i];
  for (int i = 0; i < n; i++) a[i] = m / a[i];
  int gcd = Gcd(a, n);
  return m / gcd;
}

```