#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
 
using namespace std;
 
const int MOD = 998244353;
const int MAX_VAL = 3001; // Ci 的值最大為 3000，所以 DP 狀態的第二維需要到 3001 (索引 0 到 3000)
 
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
 
    int n;
    cin >> n;
 
    vector<int> a(n);
    vector<int> b(n);
 
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    for (int i = 0; i < n; ++i) {
        cin >> b[i];
    }
 
    // dp[i][j]: 表示考慮序列 C 的前 i+1 個元素 (C_0, ..., C_i)，且 C_i = j 的合法序列數量。
    // 這裡使用 0-based 索引，所以 i 從 0 到 N-1
    vector<vector<int>> dp(n, vector<int>(MAX_VAL, 0));
 
    // 基底 (i = 0, 對應序列 C 的第一個元素 C_0)
    // C_0 可以取 A[0] 到 B[0] 之間的值
    for (int j = a[0]; j <= b[0]; ++j) {
        dp[0][j] = 1;
    }
 
    // 迭代計算 (i 從 1 到 N-1, 對應序列 C 的第 2 到 N 個元素)
    for (int i = 1; i < n; ++i) {
        // 計算前一層 dp[i-1] 的前綴和
        vector<int> prefix_sum(MAX_VAL, 0);
        prefix_sum[0] = dp[i - 1][0];
        for (int j = 1; j < MAX_VAL; ++j) {
            prefix_sum[j] = (prefix_sum[j - 1] + dp[i - 1][j]) % MOD;
        }
 
        // 計算當前層 dp[i]
        // 當前元素 C_i 可以取 A[i] 到 B[i] 之間的值 j
        for (int j = a[i]; j <= b[i]; ++j) {
            // C_{i-1} 的值 k 必須滿足 A[i-1] <= k <= B[i-1] 且 k <= C_i = j
            // 所以 k 的範圍是 [A[i-1], min(B[i-1], j)]
 
            int upper_bound_prev = min(b[i - 1], j);
            int lower_bound_prev = a[i - 1];
 
            long long count_prev = 0;
             if (lower_bound_prev <= upper_bound_prev) {
                 // 使用前綴和計算 dp[i-1] 在範圍 [lower_bound_prev, upper_bound_prev] 的和
                long long sum_up_to_upper = prefix_sum[upper_bound_prev];
                long long sum_up_to_lower_minus_1 = 0;
                if (lower_bound_prev > 0) {
                    sum_up_to_lower_minus_1 = prefix_sum[lower_bound_prev - 1];
                }
 
                count_prev = (sum_up_to_upper - sum_up_to_lower_minus_1 + MOD) % MOD; // 處理負數結果
            } else {
                 // 這個範圍不可能有 C_{i-1} 的值，數量為 0
                 count_prev = 0;
            }
 
            dp[i][j] = count_prev;
        }
    }
 
    // 計算最終答案
    // 最終答案是考慮到 C_{N-1} 時，所有可能的 C_{N-1} 值 (A[N-1] 到 B[N-1]) 對應的合法序列數量總和
    long long final_answer = 0;
    for (int j = a[n - 1]; j <= b[n - 1]; ++j) {
        final_answer = (final_answer + dp[n - 1][j]) % MOD;
    }
 
    cout << final_answer << endl;
 
    return 0;
}

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