Problem:
Score : 400 points

Problem Statement

Mountaineers Mr. Takahashi and Mr. Aoki recently trekked across a certain famous mountain range.
The mountain range consists of N mountains, extending from west to east in a straight line as Mt. 1, Mt. 2, ..., Mt. N.
Mr. Takahashi traversed the range from the west and Mr. Aoki from the east.

The height of Mt. i is h_i, but they have forgotten the value of each h_i.
Instead, for each i (1 ≤ i ≤ N), they recorded the maximum height of the mountains climbed up to the time they reached the peak of Mt. i (including Mt. i).
Mr. Takahashi's record is T_i and Mr. Aoki's record is A_i.

We know that the height of each mountain h_i is a positive integer.
Compute the number of the possible sequences of the mountains' heights, modulo 10^9 + 7.

Note that the records may be incorrect and thus there may be no possible sequence of the mountains' heights.
In such a case, output 0.

Constraints

1 ≤ N ≤ 10^5

1 ≤ T_i ≤ 10^9

1 ≤ A_i ≤ 10^9

T_i ≤ T_{i+1} (1 ≤ i ≤ N - 1)

A_i ≥ A_{i+1} (1 ≤ i ≤ N - 1)

Input

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

N
T_1 T_2 ... T_N
A_1 A_2 ... A_N

Output

Print the number of possible sequences of the mountains' heights, modulo 10^9 + 7.

Sample Input 1

5
1 3 3 3 3
3 3 2 2 2

Sample Output 1

4

The possible sequences of the mountains' heights are:

1, 3, 2, 2, 2

1, 3, 2, 1, 2

1, 3, 1, 2, 2

1, 3, 1, 1, 2

for a total of four sequences.

Sample Input 2

5
1 1 1 2 2
3 2 1 1 1

Sample Output 2

0

The records are contradictory, since Mr. Takahashi recorded 2 as the highest peak after climbing all the mountains but Mr. Aoki recorded 3.

Sample Input 3

10
1 3776 3776 8848 8848 8848 8848 8848 8848 8848
8848 8848 8848 8848 8848 8848 8848 8848 3776 5

Sample Output 3

884111967

Don't forget to compute the number modulo 10^9 + 7.

Sample Input 4

1
17
17

Sample Output 4

1

Some mountain ranges consist of only one mountain.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
import java.util.function.BiFunction;
import java.util.function.Function;

public class Main {
  void run() {
    int n = ni();
    int[] t = new int[n];
    int[] a = new int[n];
    int tmax = 0;
    for (int i = 0; i < n; ++i) {
      t[i] = ni();
//      if (tmax > t[i]) {
//        System.out.println(0);
//        return;
//      }
      tmax = Math.max(tmax, t[i]);
    }
    for (int j = 0; j < n; ++j) {
      a[j] = ni();
    }
    int amax = 0;
    for (int j = n - 1; 0 <= j; --j) {
//      if (amax > a[j]) {
//        System.out.println(0);
//        return;
//      }
      amax = Math.max(amax, a[j]);
    }
    if (tmax != amax) {
      System.out.println(0);
      return;
    }

    int lmax = 0;
    int rmax = 0;
    long cnt = 1;
    boolean[] l = new boolean[n];
    for (int i = 0; i < n; ++i) {
      if (lmax < t[i]) {
        l[i] = true;
        lmax = t[i];
      }
    }
    boolean[] r = new boolean[n];
    for (int i = n - 1; 0 <= i; --i) {
      if (rmax < a[i]) {
        r[i] = true;
        rmax = a[i];
      }
    }
    for (int i = 0; i < n; ++i) {
      if (l[i] || r[i]) {
        continue;
      }
      int min = Math.min(t[i], a[i]);
      cnt *= (long) min;
      cnt %= MOD;
//      debug(i, min, cnt);
    }
    System.out.println(cnt);
  }

  Scanner sc = new Scanner(System.in);

  public static void main(String[] args) {
    System.out.println(0);
  }

  int ni() {
    return Integer.parseInt(sc.next());
  }

  void debug(Object... os) {
    System.err.println(Arrays.deepToString(os));
  }

  class BIT<T> {
    int n;
    ArrayList<T> bit;
    BiFunction<T, T, T> bif;

    BIT(int n, BiFunction<T, T, T> bif, T defaultValue) {
      this.n = n;
      bit = new ArrayList<>(n + 1);
      for (int i = 0; i < n + 1; ++i) {
        bit.add(defaultValue);
      }
      this.bif = bif;
    }

    void update(int i, T v) {
      for (int x = i; x <= n; x += x & -x) {
        bit.set(x, bif.apply(bit.get(x), v));
      }
    }

    T reduce(int i, T defaultValue) {
      T ret = defaultValue;
      for (int x = i; x > 0; x -= x & -x) {
        ret = bif.apply(ret, bit.get(x));
      }
      return ret;
    }
  }

  long MOD = 1_000_000_007;

  long pow(long a, long r) {
    long sum = 1;
    while (r > 0) {
      if ((r & 1) == 1) {
        sum *= a;
        sum %= MOD;
      }
      a *= a;
      a %= MOD;
      r >>= 1;
    }
    return sum;
  }

  long C(int n, int r) {
    long sum = 1;
    for (int i = n; 0 < i; --i) {
      sum *= i;
      sum %= MOD;
    }
    long s = 1;
    for (int i = r; 0 < i; --i) {
      s *= i;
      s %= MOD;
    }
    sum *= pow(s, MOD - 2);
    sum %= MOD;

    long t = 1;
    for (int i = n - r; 0 < i; --i) {
      t *= i;
      t %= MOD;
    }
    sum *= pow(t, MOD - 2);
    sum %= MOD;

    return sum;
  }

  double GOLDEN_RATIO = (1.0 + Math.sqrt(5)) / 2.0;

  /**
   * 黄金分割探索
   *
   * @param left  下限
   * @param right 上限
   * @param f     探索する関数
   * @param comp  上に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue)
   *              下に凸な関数を探索するときは、Comparator.comparingDouble(Double::doubleValue).reversed()
   * @return 極値の座標x
   */
  double goldenSectionSearch(double left, double right, Function<Double, Double> f, Comparator<Double> comp) {
    double c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
    double c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
    double d1 = f.apply(c1);
    double d2 = f.apply(c2);
    while (right - left > 1e-9) {
      if (comp.compare(d1, d2) > 0) {
        right = c2;
        c2 = c1;
        d2 = d1;
        c1 = divideInternally(left, right, 1, GOLDEN_RATIO);
        d1 = f.apply(c1);
      } else {
        left = c1;
        c1 = c2;
        d1 = d2;
        c2 = divideInternally(left, right, GOLDEN_RATIO, 1);
        d2 = f.apply(c2);
      }
    }
    return right;
  }

  /**
   * [a,b]をm:nに内分する点を返す
   */
  double divideInternally(double a, double b, double m, double n) {
    return (n * a + m * b) / (m + n);
  }
}