Problem:
Score : 1400 points

Problem Statement

Snuke is having another barbeque party.

This time, he will make one serving of Skewer Meal.

He has a stock of N Skewer Meal Packs. The i-th Skewer Meal Pack contains one skewer, A_i pieces of beef and B_i pieces of green pepper.
All skewers in these packs are different and distinguishable, while all pieces of beef and all pieces of green pepper are, respectively, indistinguishable.

To make a Skewer Meal, he chooses two of his Skewer Meal Packs, and takes out all of the contents from the chosen packs, that is, two skewers and some pieces of beef or green pepper.
(Remaining Skewer Meal Packs will not be used.)
Then, all those pieces of food are threaded onto both skewers, one by one, in any order.

(See the image in the Sample section for better understanding.)

In how many different ways can he make a Skewer Meal? Two ways of making a Skewer Meal is different if and only if the sets of the used skewers are different, or the orders of the pieces of food are different.
Since this number can be extremely large, find it modulo 10^9+7.

Constraints

2≦N≦200,000

1≦A_i≦2000, 1≦B_i≦2000

Input

The input is given from Standard Input in the following format:

N
A_1 B_1
A_2 B_2
:
A_N B_N

Output

Print the number of the different ways Snuke can make a serving of Skewer Meal, modulo 10^9+7.

Sample Input 1

3
1 1
1 1
2 1

Sample Output 1

26

The 26 ways of making a Skewer Meal are shown below.
Gray bars represent skewers, each with a number denoting the Skewer Meal Set that contained the skewer.
Brown and green rectangles represent pieces of beef and green pepper, respectively.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

// Copyright (C) 2017 __debug.

// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.


#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/priority_queue.hpp>

#define x first
#define y second
#define MP std::make_pair
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(), (x).end()
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#ifdef __linux__
#define getchar getchar_unlocked
#define putchar putchar_unlocked
#endif

using std::pair;
using std::vector;
using std::string;

typedef long long LL;
typedef pair<int, int> Pii;

const int oo = 0x3f3f3f3f;

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, true : false; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, true : false; }
string procStatus()
{
    std::ifstream t("/proc/self/status");
    return string(std::istreambuf_iterator<char>(t), std::istreambuf_iterator<char>());
}
template<typename T> T read(T &x)
{
    int f = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
        f = (ch == '-' ? -1 : 1);
    for (x = 0; isdigit(ch); ch = getchar())
        x = 10 * x + ch - '0';
    return x *= f;
}
template<typename T> void write(T x)
{
    if (x == 0) {
        putchar('0');
        return;
    }
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    static char s[20];
    int top = 0;
    for (; x; x /= 10)
        s[++top] = x % 10 + '0';
    while (top)
        putchar(s[top--]);
}
// EOT

const int MAXN = 2e5 + 5, RNG = 2e3 + 5;
const int MOD = 1e9 + 7;

int N;
Pii A[MAXN];

void input()
{
    read(N);
    for (int i = 1; i <= N; ++i) {
        read(A[i].x); read(A[i].y);
    }
}

int fac[RNG * 4], ifac[RNG * 4];

LL fpm(LL base, LL exp)
{
    LL ret = 1;
    for (; exp; exp >>= 1) {
        if (exp & 1)
            (ret *= base) %= MOD;
        (base *= base) %= MOD;
    }
    return ret;
}

inline int C(int n, int k)
{
    assert(0 <= k && k <= n);
    return (LL)fac[n] * ifac[k] % MOD * ifac[n - k] % MOD;
}

void solve()
{
    fac[0] = 1;
    for (int i = 1; i < RNG * 4; ++i) {
        fac[i] = (LL)fac[i - 1] * i % MOD;
    }
    ifac[RNG * 4 - 1] = fpm(fac[RNG * 4 - 1], MOD - 2);
    for (int i = RNG * 4 - 1; i >= 1; --i) {
        ifac[i - 1] = (LL)ifac[i] * i % MOD;
    }

    static int dp[RNG * 2][RNG * 2];
    static int cnt[RNG * 2][RNG * 2];
    int ans = 0;

    for (int i = 1; i <= N; ++i) {
        dp[RNG - A[i].x][RNG - A[i].y] += 1;
        cnt[RNG + A[i].x][RNG + A[i].y] += 1;
        (ans += MOD - C(A[i].x * 2 + A[i].y * 2, A[i].x * 2)) %= MOD;
    }
    for (int i = 0; i < RNG * 2; ++i) {
        for (int j = 0; j < RNG * 2; ++j) {
            (dp[i][j] += (i ? dp[i - 1][j] : 0) + (j ? dp[i][j - 1] : 0)) %= MOD;
            (ans += (LL)cnt[i][j] * dp[i][j] % MOD) %= MOD;
        }
    }
    ans = (LL)ans * fpm(2, MOD - 2) % MOD;

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

int main()
{
#ifdef __DEBUG
    freopen("E.in", "r", stdin);
    freopen("E.out", "w", stdout);
#endif

    input();
    solve();

    return 0;
}

// 念兰堂红烛，心长焰短，向人垂泪。
//     -- 晏殊《撼庭秋·别来音信千里》