Problem:
Score : 300 points

Problem Statement

You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N.
Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to a_i.

Add 1 to b_j.

Constraints

1 ≤ N ≤ 10 000

0 ≤ a_i,b_i ≤ 10^9 (1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N
b_1 b_2 .. b_N

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

3
1 2 3
5 2 2

Sample Output 1

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a = \{3,2,3\}, b = \{5,3,2\}.

Second operation: i=1 and j=2. Now we have a = \{5,2,3\}, b = \{5,4,2\}.

Third operation: i=2 and j=3. Now we have a = \{5,4,3\}, b = \{5,4,3\}.

Sample Input 2

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

No

Sample Input 3

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

No

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

#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <list>
#include <vector>
#include <complex>
#include <utility>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <climits>
#include <bitset>
#include <ctime>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <cassert>
#include <cstddef>
#include <iomanip>
#include <numeric>
#include <tuple>
#include <sstream>
#include <fstream>

using namespace std;
#define REP(i, n) for (int (i) = 0; (i) < (n); (i)++)
#define FOR(i, a, b) for (int (i) = (a); (i) < (b); (i)++)
#define RREP(i, a) for(int (i) = (a) - 1; (i) >= 0; (i)--)
#define FORR(i, a, b) for(int (i) = (a) - 1; (i) >= (b); (i)--)
#define DEBUG(C) cerr << #C << " = " << C << endl;
using LL = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<LL>;
using VVL = vector<VL>;
using VD = vector<double>;
using VVD = vector<VD>;
using PII = pair<int, int>;
using PDD = pair<double, double>;
using PLL = pair<LL, LL>;
using VPII = vector<PII>;
template<typename T> using VT = vector<T>;
#define ALL(a) begin((a)), end((a))
#define RALL(a) rbegin((a)), rend((a))
#define SORT(a) sort(ALL((a)))
#define RSORT(a) sort(RALL((a)))
#define REVERSE(a) reverse(ALL((a)))
#define MP make_pair
#define FORE(a, b) for (auto &&a : (b))
#define FIND(s, e) ((s).find(e) != (s).end())
#define EB emplace_back
template<typename T> inline bool chmax(T &a, T b){if (a < b){a = b;return true;}return false;}
template<typename T> inline bool chmin(T &a, T b){if (a > b){a = b;return true;}return false;}

const int INF = 1e9;
const int MOD = INF + 7;
const LL LLINF = 1e18;

const int MAX = 1e5 + 10;
int n;
int a[MAX], b[MAX];

int main(void) {
    cin >> n;
    REP(i, n) cin >> a[i];
    REP(i, n) cin >> b[i];
    VI v;
    REP(i, n) {
        v.EB(a[i] - b[i]);
    }
    FORE(e, v) cerr << e << " ";cerr << endl;
    int must = 0;
    VI minus;
    FORE(e, v) {
        if (e > 0) must += e;
        else if (e < 0) minus.EB(e);
    }
    must *= 2;
    FORE(e, minus) {
        must += e;
    }
    DEBUG(must)
    cout << (must <= 0 ? "Yes" : "No") << endl;
}