#include <bits/stdc++.h>
using namespace std;
template <typename T>
void inline chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void inline chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
template <typename T>
void print(T x) {
  if (x < 0) {
    putchar('-');
    print(x);
    return;
  }
  if (x >= 10) print(x / 10);
  putchar((x % 10) + '0');
}
const int N = 105;
int n, ans;
int c[N], a[N], b[N];
int main() {
  read(n);
  for (int i = 1; i <= n; i++) scanf("%1d", c + i);
  for (int i = 1; i <= n; i++) scanf("%d%d", a + i, b + i);
  for (int t = 0; t <= 10000; t++) {
    int s = 0;
    for (int i = 1; i <= n; i++) {
      if (t < b[i]) {
        if (c[i] == 1) s++;
      } else {
        int st = c[i] ^ (((t - b[i]) / a[i] + 1) & 1);
        if (st == 1) s++;
      }
    }
    chkMax(ans, s);
  }
  printf("%d\n", ans);
}

### Prompt

Generate a cpp solution to 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;
template <typename T>
void inline chkMax(T &x, T y) {
  if (y > x) x = y;
}
template <typename T>
void inline chkMin(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
void inline read(T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s < '0' || s > '9') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s <= '9' && s >= '0') x = x * 10 + (s ^ 48), s = getchar();
  x *= f;
}
template <typename T>
void print(T x) {
  if (x < 0) {
    putchar('-');
    print(x);
    return;
  }
  if (x >= 10) print(x / 10);
  putchar((x % 10) + '0');
}
const int N = 105;
int n, ans;
int c[N], a[N], b[N];
int main() {
  read(n);
  for (int i = 1; i <= n; i++) scanf("%1d", c + i);
  for (int i = 1; i <= n; i++) scanf("%d%d", a + i, b + i);
  for (int t = 0; t <= 10000; t++) {
    int s = 0;
    for (int i = 1; i <= n; i++) {
      if (t < b[i]) {
        if (c[i] == 1) s++;
      } else {
        int st = c[i] ^ (((t - b[i]) / a[i] + 1) & 1);
        if (st == 1) s++;
      }
    }
    chkMax(ans, s);
  }
  printf("%d\n", ans);
}

```