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<iostream>

using namespace std;
int main(){
int N,M,s[5],c[5],i,a=0;
int X[4];
X[1]=-1;
X[2]=-1;
X[3]=-1;

cin >> N >> M;
for(i=1;i<=M;i++){cin >> s[i] >> c[i];}

for(i=1;i<=M;i++){if((X[s[i]]==-1)||(X[s[i]]==c[i])){X[s[i]]=c[i];}else{a--;
break;}
    if(s[i]==1&&c[i]==0&&N!=1){a--;}
}
if(a==0){for(i=1;i<=N;i++){
if(i==1&&X[i]==-1&&N!=1){X[i]=1;}
if(i==1&&X[i]==-1&&N==1){X[i]=0;}
if((i==2||i==3)&&X[i]==-1){X[i]=0;}
}
}
if(a<0){cout << "-1";}else{
for(i=1;i<=N;i++){cout << X[i];}}
cout << endl;
return 0;
}