1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <vector> 5 #include<set> 6 #include <map> 7 #include <string> 8 using namespace std; 9 10 #define ll long long 11 #define pb push_back 12 13 const ll MOD = 19940417; 14 const int N = 1e6; 15 16 17 ll inv2, inv6; 18 19 // ll gcd( ll a, ll b) 20 // { 21 // return b == 0 ? a : gcd(b, a % b); 22 // } 23 24 ll ksm(ll a, ll b, ll mod) 25 { 26 ll res = 1; 27 while(b){ 28 if(b & 1) res = (res * a) % mod; 29 a = a * a % mod; 30 b >>= 1; 31 } 32 return res; 33 } 34 35 inline ll fx2(ll n) 36 { 37 return n * (n + 1) % MOD * inv2 % MOD; 38 } 39 40 ll fun1(ll n) 41 { 42 ll l, r, sum1, sum2; 43 sum1 = sum2 = 0; 44 sum1 = n * n % MOD; 45 for(l = 1, r = 0; l <= n; l = r + 1){ 46 r = n / (n / l); 47 sum2 = (sum2 + (fx2(r) - fx2(l - 1) + MOD) * (n / l) % MOD) % MOD; 48 } 49 return (sum1 - sum2 + MOD) % MOD; 50 } 51 52 ll fun2(ll n, ll m) 53 { 54 ll l, r, sum1, sum2; 55 sum1 = sum2 = 0; 56 //cout << n << " " << m << endl; 57 for(l = 1, r = 0; l <= n; l = r + 1){ 58 r = n / (n / l); 59 sum1 = (sum1 + (fx2(r) - fx2(l - 1) + MOD) * (n / l)) % MOD; 60 } 61 sum1 = m % MOD * sum1 % MOD; 62 63 for(l = 1, r = 0; l <= n; l = r + 1){ 64 r = min(n, m / (m / l)); 65 sum2 = (sum2 + (fx2(r) - fx2(l - 1) + MOD) * (m / l)) % MOD; 66 } 67 sum2 = n % MOD * sum2 % MOD; 68 69 return (sum1 + sum2) % MOD; 70 } 71 72 inline ll fx6(ll n) 73 { 74 return n * (n + 1) % MOD * (2 * n + 1) % MOD * inv6 % MOD; 75 } 76 77 ll fun3(ll n, ll m) 78 { 79 ll l, r, sum; 80 sum = 0; 81 for(l = 1, r = 0; l <= n; l = r + 1){ 82 //取右边小的使得值正确,如果取到了右边大的,会导致a或者b变小 83 r = min(n / (n / l), m / (m / l)); 84 // /r = n / (n / l); 85 ll a = n / l; 86 ll b = m / l; 87 sum = (sum + (fx6(r) - fx6(l - 1) + MOD) * a % MOD * b % MOD) % MOD; 88 } 89 return sum % MOD; 90 } 91 92 93 94 ll exgcd(ll a, ll b, ll &x, ll &y) 95 { 96 97 if(b == 0){//推理1,终止条件 98 x = 1; 99 y = 0; 100 return a; 101 } 102 ll r = exgcd(b, a%b, x, y); 103 //先得到更底层的x2,y2,再根据计算好的x2,y2计算x1,y1。 104 //推理2,递推关系 105 ll t = y; 106 y = x - (a/b) * y; 107 x = t; 108 return r; 109 } 110 111 void inv() 112 { 113 114 //ax同余1(mod prime) 求最小正整数解 115 // ax + by = n ---> ax' + by' = gcd(a, b) 116 // k = gcd(a, b) k*(ax' + by') = n 117 //以下 a, b, x, y对应上面 a, b, x', y' 118 ll x, y; 119 //以下求出mod不为质数的逆元 120 // a / b % mod c 121 // a / b * b^-1 ≡ x * b^-1(mod c) 122 // b * b^-1 ≡ 1 (mod c) 123 // b * b^-1 + y * c = 1 124 // a * x + b * y = 1 125 exgcd(2, MOD, x, y); 126 inv2 = (x + MOD) % MOD; 127 exgcd(6, MOD, x, y); 128 inv6 = (x + MOD) % MOD; 129 130 // cout << "inv2 = " << (inv2 + MOD) % MOD << endl; 131 // cout << "inv6 = " << (inv6 + MOD) % MOD << endl; 132 } 133 134 void solve() 135 { 136 137 //cout << gcd(2, MOD) << " " << gcd(6, MOD) << endl; 138 //cout << inv2 << " " << inv6 << endl; 139 inv(); 140 // cout << inv2 << " " << inv6 << endl; 141 ll n, m; 142 cin >> n >> m; 143 if(n > m) swap(n, m); 144 ll ans1 = (fun1(n) * fun1(m)) % MOD; 145 ll ans2 = (n * n % MOD * m % MOD - fun2(n, m) + fun3(n, m) + MOD) % MOD; 146 ll ans = (ans1 - ans2 + MOD) % MOD; 147 148 cout << ans << endl; 149 } 150 151 int main() 152 { 153 ios::sync_with_stdio(false); 154 cin.tie(0); 155 cout.tie(0); 156 solve(); 157 //int ok = 0; 158 //cout << "ok" << endl; 159 return 0; 160 }