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
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(v) v.begin(), v.end()
using namespace std;
typedef long long ll;

int main() {
	int n, m;
	cin >> n >> m;
	vector<int> s(m), c(m);
	rep(i, m) cin >> s[i] >> c[i];
	int i = pow(10, n-1);
	if(n == 1) i = 0;
	for( ; i < pow(10, n); i++) {
		bool ok = true;
		rep(j, m) {
			if((i/(int)pow(10, n-s[j]))%10 != c[j]) {
				ok = false;
				break;
			}
		}
		if(ok) {
			cout << i << endl;
			return 0;
		}
	}
	cout << -1 << endl;
	
	return 0;
}