Problem: There are r red and g green blocks for construction of the red-green tower. Red-green tower can be built following next rules:

  * Red-green tower is consisting of some number of levels;

  * Let the red-green tower consist of n levels, then the first level of this tower should consist of n blocks, second level — of n - 1 blocks, the third one — of n - 2 blocks, and so on — the last level of such tower should consist of the one block. In other words, each successive level should contain one block less than the previous one;

  * Each level of the red-green tower should contain blocks of the same color.

<image>

Let h be the maximum possible number of levels of red-green tower, that can be built out of r red and g green blocks meeting the rules above. The task is to determine how many different red-green towers having h levels can be built out of the available blocks.

Two red-green towers are considered different if there exists some level, that consists of red blocks in the one tower and consists of green blocks in the other tower.

You are to write a program that will find the number of different red-green towers of height h modulo 109 + 7.

Input

The only line of input contains two integers r and g, separated by a single space — the number of available red and green blocks respectively (0 ≤ r, g ≤ 2·105, r + g ≥ 1).

Output

Output the only integer — the number of different possible red-green towers of height h modulo 109 + 7.

Examples

Input

4 6


Output

2


Input

9 7


Output

6


Input

1 1


Output

2

Note

The image in the problem statement shows all possible red-green towers for the first sample.

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
public class RedGreenTowers {
    
    
    
    /************************ SOLUTION STARTS HERE ************************/
    
    static final int mod = (int) 1e9 + 7;

    private static void solve() {
        
        int R = nextInt();
        int G = nextInt();
        
        int DPprev[] = new int[R + 1];  // number of ways of building blocks using exactly i red blocks
        DPprev[0] = G > 0 ? 1 : 0;
        
        if(R > 0)
            DPprev[1] = 1;
        
        boolean flag = false;
        int h = 2;
        int hMax = (int) Math.ceil((Math.sqrt(1 + 8 * (R + G)) - 1) / 2.0) + 1;
        for(int run = 1; run <= hMax; run++) {   
            int DPnew[] = new int[R + 1];
            flag = false;
            for(int i = 0; i <= R; i++) {
                int j = ((h * (h + 1)) / 2) - i;    // green
                if(i >= h) {    // build base with red
                    DPnew[i] = DPprev[i - h];
                }
                if(j >= h && j <= G) {    // build base with green
                    DPnew[i] = (DPnew[i] + DPprev[i]) % mod;
                }
                
                flag |= DPnew[i] != 0;
            }
            
            if(!flag)
                break;

            DPprev = DPnew;
            h++;
            
        }
        int sum = 0;
        for(int a : DPprev)
            sum = (sum + a) % mod;
            
        println(sum);
        
    }
    
    
    
    /************************ SOLUTION ENDS HERE ************************/
    
    
    
    
    
    /************************ TEMPLATE STARTS HERE **********************/
    
    public static void main(String[] args) throws IOException {
        reader = new BufferedReader(new InputStreamReader(System.in));
        writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
        st     = null;
        solve();
        reader.close();
        writer.close();
    }
    
    static BufferedReader reader;
    static PrintWriter    writer;
    static StringTokenizer st;
    
    static String next()
    {while(st == null || !st.hasMoreTokens()){try{String line = reader.readLine();if(line == null){return null;}            
    st = new StringTokenizer(line);}catch (Exception e){throw new RuntimeException();}}return st.nextToken();}
    static String nextLine()  {String s=null;try{s=reader.readLine();}catch(IOException e){e.printStackTrace();}return s;}             
    static int    nextInt()   {return Integer.parseInt(next());}
    static long   nextLong()  {return Long.parseLong(next());}     
    static double nextDouble(){return Double.parseDouble(next());}
    static char   nextChar()  {return next().charAt(0);}
    static int[]  nextIntArray(int n)         {int[] a= new int[n];   int i=0;while(i<n){a[i++]=nextInt();}  return a;}
    static long[] nextLongArray(int n)        {long[]a= new long[n];  int i=0;while(i<n){a[i++]=nextLong();} return a;}    
    static int[]  nextIntArrayOneBased(int n) {int[] a= new int[n+1]; int i=1;while(i<=n){a[i++]=nextInt();} return a;}            
    static long[] nextLongArrayOneBased(int n){long[]a= new long[n+1];int i=1;while(i<=n){a[i++]=nextLong();}return a;}            
    static void   print(Object o)  { writer.print(o);  }
    static void   println(Object o){ writer.println(o);}
    
    /************************ TEMPLATE ENDS HERE ************************/
    
}