If there is an integer not less than 0 satisfying the following conditions, print the smallest such integer; otherwise, print `-1`.

* The integer has exactly N digits in base ten. (We assume 0 to be a 1-digit integer. For other integers, leading zeros are not allowed.)
* The s_i-th digit from the left is c_i. \left(i = 1, 2, \cdots, M\right)

Constraints

* All values in input are integers.
* 1 \leq N \leq 3
* 0 \leq M \leq 5
* 1 \leq s_i \leq N
* 0 \leq c_i \leq 9

Input

Input is given from Standard Input in the following format:


N M
s_1 c_1
\vdots
s_M c_M


Output

Print the answer.

Examples

Input

3 3
1 7
3 2
1 7


Output

702


Input

3 2
2 1
2 3


Output

-1


Input

3 1
1 0


Output

-1
/*input
1 1
1 0
*/
#include <bits/stdc++.h>
using namespace std;
#define MAX 5

int a[MAX] = {}, vis[MAX] = {};
int n, m;

int main() {
	cin >> n >> m;
	for (int i = 1; i <= m; i++) {
		int p, x; cin >> p >> x;

		if (vis[p] && a[p] != x) {
			cout << -1 << '\n';
			return 0;
		}
		if (p == 1 && x == 0 && n != 1) {
			cout << -1 << '\n';
			return 0;
		}

		vis[p] = 1;
		a[p] = x;
	}
	if (!vis[1] && n != 1) a[1] = 1;

	for (int i = 1; i <= n; i++) cout << a[i];
	cout << '\n';
	return 0;
}