#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
using namespace std;

typedef long long ll;
const int MOD = (int)1e9 + 7;
int add(int x, int y)
{
    x += y;
    if (x >= MOD) return x - MOD;
    return x;
}
int sub(int x, int y)
{
    x -= y;
    if (x < 0) return x + MOD;
    return x;
}
int mult(int x, int y)
{
    return ((ll)x * y) % MOD;
}
int bin_pow(int x, int p)
{
    if (p == 0) return 1;
    if (p == 2 || (p & 1)) return mult(x, bin_pow(x, p - 1));
    return bin_pow(bin_pow(x, p / 2), 2);
}
int rev(int x)
{
    return bin_pow(x, MOD - 2);
}

const int N = 200100;
const int K = 4010;
const int Z = 2002;
int n;
int a[N][2];
int C[K][K];
ll b[K];
int f[2 * K], rf[2 * K];
int ans = 0;

void init()
{
    f[0] = 1;
    for (int i = 1; i < 2 * K; i++)
        f[i] = mult(f[i - 1], i);
    rf[2 * K - 1] = rev(f[2 * K - 1]);
    for (int i = 2 * K - 1; i > 0; i--)
        rf[i - 1] = mult(rf[i], i);


    for (int i = 0; i < K; i++)
        C[i][0] = C[i][i] = 1;
    for (int i = 1; i < K; i++)
        for (int j = 1; j < i; j++)
            C[i][j] = add(C[i - 1][j], C[i - 1][j - 1]);

    return;
}

void read()
{
/*
    n = 200000;
    for (int i = 0; i < n; i++)
        a[i][0] = a[i][1] = 2000;
    */


    scanf("%d", &n);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < 2; j++)
            scanf("%d", &a[i][j]);


    return;
}

void solve(int x, int y)
{
    /*
    for (int i = -x; i <= y; i++)
        b[Z - i] = add(b[Z - i], C[x + y][x + i]);
    */

    for (int i = -x; i <= y; i++)
        b[Z - i] += C[x + y][x + i];

    return;
}

int getC(int n, int k)
{
    return mult(f[n], mult(rf[k], rf[n - k]));
}

int main()
{
//  freopen("input.txt", "r", stdin);
//  freopen("output.txt", "w", stdout);

    init();

    read();

    for (int i = 0; i < n; i++)
        solve(a[i][0], a[i][1]);

    for (int i = -Z; i <= Z; i++)
        b[Z + i] %= MOD;

    for (int i = -Z; i <= Z; i++)
        ans = add(ans, mult(b[Z + i], b[Z - i]));

    for (int i = 0; i < n; i++)
        ans = sub(ans, getC(2 * (a[i][0] + a[i][1]), 2 * a[i][0]));

    if (ans & 1) ans += MOD;
    ans /= 2;

    printf("%d\n", ans);

    return 0;
}