## Algorithm

B. Hamster Farm
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Dima has a hamsters farm. Soon N hamsters will grow up on it and Dima will sell them in a city nearby.

Hamsters should be transported in boxes. If some box is not completely full, the hamsters in it are bored, that's why each box should be completely full with hamsters.

Dima can buy boxes at a factory. The factory produces boxes of K kinds, boxes of the i-th kind can contain in themselves ai hamsters. Dima can buy any amount of boxes, but he should buy boxes of only one kind to get a wholesale discount.

Of course, Dima would buy boxes in such a way that each box can be completely filled with hamsters and transported to the city. If there is no place for some hamsters, Dima will leave them on the farm.

Find out how many boxes and of which type should Dima buy to transport maximum number of hamsters.

Input

The first line contains two integers N and K (0 ≤ N ≤ 10181 ≤ K ≤ 105) — the number of hamsters that will grow up on Dima's farm and the number of types of boxes that the factory produces.

The second line contains K integers a1a2, ..., aK (1 ≤ ai ≤ 1018 for all i) — the capacities of boxes.

Output

Output two integers: the type of boxes that Dima should buy and the number of boxes of that type Dima should buy. Types of boxes are numbered from 1 to K in the order they are given in input.

If there are many correct answers, output any of them.

Examples
input
Copy
`19 35 4 10`
output
Copy
`2 4`
input
Copy
`28 35 6 30`
output
Copy
`1 5`

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>

using namespace std;

int const N = 1e5 + 1;
long long n, k, a[N];

int main() {
scanf("%lld %lld", &n, &k);
for(int i = 0; i  <  k; ++i)
scanf("%lld", a + i);

if(n == 0) {
cout << 1 << ' ' << 0 << endl;
return 0;
}

long long mn = 2e18, idx = k - 1;
for(int i = 0; i < k; ++i) {
if(a[i]  < = n && n % a[i] < mn> {
mn = n % a[i];
idx = i;
}
}

cout << idx + 1 << ' ' << n / a[idx] << endl;

return 0;
}``````
Copy The Code &

Input

cmd
19 3
5 4 10

Output

cmd
2 4