This morning Tolik has understood that while he was sleeping he had invented an incredible problem which will be a perfect fit for Codeforces! But, as a "Discuss tasks" project hasn't been born yet (in English, well), he decides to test a problem and asks his uncle.

After a long time thinking, Tolik's uncle hasn't any ideas on how to solve it. But, he doesn't want to tell Tolik about his inability to solve it, so he hasn't found anything better than asking you how to solve this task.

In this task you are given a cell field n ⋅ m, consisting of n rows and m columns, where point's coordinates (x, y) mean it is situated in the x-th row and y-th column, considering numeration from one (1 ≤ x ≤ n, 1 ≤ y ≤ m). Initially, you stand in the cell (1, 1). Every move you can jump from cell (x, y), which you stand in, by any non-zero vector (dx, dy), thus you will stand in the (x+dx, y+dy) cell. Obviously, you can't leave the field, but also there is one more important condition — you're not allowed to use one vector twice. Your task is to visit each cell of the field exactly once (the initial cell is considered as already visited).

Tolik's uncle is a very respectful person. Help him to solve this task!

Input

The first and only line contains two positive integers n, m (1 ≤ n ⋅ m ≤ 10^{6}) — the number of rows and columns of the field respectively.

Output

Print "-1" (without quotes) if it is impossible to visit every cell exactly once.

Else print n ⋅ m pairs of integers, i-th from them should contain two integers x_i, y_i (1 ≤ x_i ≤ n, 1 ≤ y_i ≤ m) — cells of the field in order of visiting, so that all of them are distinct and vectors of jumps between them are distinct too.

Notice that the first cell should have (1, 1) coordinates, according to the statement.

Examples

Input


2 3


Output


1 1
1 3
1 2
2 2
2 3
2 1

Input


1 1


Output


1 1

Note

The vectors from the first example in the order of making jumps are (0, 2), (0, -1), (1, 0), (0, 1), (0, -2).
import java.io.*; 
import java.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
import javax.swing.*;
import java.awt.*;
 
public class tempo 
{ 
    static class Pair{
        int x;int y;int i;
        Pair(int x,int y,int i){
            this.x=x;
            this.y=y;
            this.i=i;
        }
    }
    static class sorting implements Comparator<Pair>{
        public int compare(Pair a,Pair b){
            //return (a.y)-(b.y);
            return (a.x-b.x);
        }
    }
 
     static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 
    
 
   // static int n;static ArrayList<Pair> arr1;
    //static int len;
    public static int[] na(int n)throws IOException{
        Reader s1=new Reader();
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = s1.nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    public static void main(String[] args) throws IOException { 
        Reader s=new Reader();
       // Scanner s=new Scanner(System.in);
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int n=s.nextInt();
        int m=s.nextInt();
        int x1=1, y1=1, x2=n, y2=m;
        for (int i=0; i<n*m; i++) {
            if (i%2 == 0) {
                pw.println(x1+" "+y1);
             //   System.out.println("yes");
                x1++;
                if (x1 > n) {
                    x1 = 1;
                    y1++;
                }
            }  else {
                pw.println(x2+" "+y2);
                x2--;
                if (x2 < 1) {
                    x2 = n;
                    y2--;
                }
            }
        }
        pw.close();
    }
   
  
    
// Returns factorial of n 
static int fact(int n) 
{ 
    int res = 1; 
    for (int i = 2; i <= n; i++) 
        res = res * i; 
    return res; 
}
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static long gcd(long a, long b) 
    { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
        static long modInverse(long a, long m) 
    { 
        long g = gcd(a, m); 
        if (g != 1) 
            return -1; 
        else 
        { 
            // If a and m are relatively prime, then modulo inverse 
            // is a^(m-2) mode m 
            return (power(a, m - 2, m)); 
        } 
    } 
      
    // To compute x^y under modulo m 
    static long power(long x, long y, long m)  
    { 
        if (y == 0) 
            return 1; 
          
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
     }
    // static long fast_pow(long base,long n,long M)
    // {
    //     if(n==0)
    //        return 1;
    //     if(n==1)
    //     return base;
    //     long halfn=fast_pow(base,n/2,M);
    //     if(n%2==0)
    //         return ( halfn * halfn ) % M;
    //     else
    //         return ( ( ( halfn * halfn ) % M ) * base ) % M;
    // }
    //    static long modInverse(long n,int M)
    // {
    //     return fast_pow(n,M-2,M);
    // }
    
} 