#include #include #define rep(i,b) for(int i=0;i=0;i--) #define rep1(i,b) for(int i=1;i=x;i--) #define fore(i,a) for(auto i:a) #define fore1(i,a) for(auto &i:a) #define rng(x) (x).begin(), (x).end() #define rrng(x) (x).rbegin(), (x).rend() #define sz(x) ((int)(x).size()) #define pb push_back #define fi first #define se second using namespace std; using namespace atcoder; using ll = long long; template using mpq = priority_queue, greater>; template bool chmax(T &a, const T &b) { if (a bool chmin(T &a, const T &b) { if (b ll sumv(const vector&a){ll res(0);for(auto&&x:a)res+=x;return res;} bool yn(bool a) { if(a) {cout << "Yes" << endl; return 1;} else {cout << "No" << endl; return 0;}} #define dame { cout << "No" << endl; return 0;} #define dame1 { cout << -1 << endl; return 0;} #define test(x) cout << "test" << x << endl; #define deb(x,y) cout << x << " " << y << endl; #define deb3(x,y,z) cout << x << " " << y << " " << z << endl; #define deb4(x,y,z,x2) cout << x << " " << y << " " << z << " " << x2 << endl; #define out cout << ans << endl; #define outv fore(yans , ans) cout << yans << "\n"; #define show(x) cerr<<#x<<" = "<; using pil = pair; using pli = pair; using pii = pair; using tp = tuple; using vi = vector; using vl = vector; using vs = vector; using vb = vector; using vpii = vector; using vpli = vector; using vpll = vector; using vpil = vector; using vvi = vector>; using vvl = vector>; using vvs = vector>; using vvb = vector>; using vvpii = vector>; using vvpli = vector>; using vvpll = vector; using vvpil = vector; using mint = modint998244353; using vm = vector; using vvm = vector>; vector dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1}; ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;} ll lcm(ll a, ll b) { return a/gcd(a,b)*b;} const double eps = 1e-10; const ll LINF = 1001002003004005006ll; const int INF = 1001001001; int main(){ ll l,r; cin>>l>>r; ll a,b,c; cin>>a>>b>>c; mint ans = 0; if (a+b >= c){ ll t = (r-l+1 - (a-1) - (b-1)); if (t < 3) ans = 0; else ans = mint(t) * mint(t-1) * mint(t-2) / 6; }else{ if (c > r-l) {cout << 0 << endl; return 0;} mint tmp2 = 0; mint tmp1 = 0; mint tmp0 = 0; tmp1 += mint(r-l) * (r-l+1) / 2; tmp1 -= mint(c) * (c-1) / 2; tmp1 *= mint(r-l+a+b); tmp0 -= mint(r-l - c + 1) * mint(b+a-1) * mint(r-l+1); mint rr = r-l; tmp2 -= rr * (2*rr+1) * (rr + 1) / 6; rr = c-1; tmp2 += rr * (2*rr+1) * (rr + 1) / 6; ans = tmp2 + tmp1 + tmp0; } cout << ans.val() << endl; return 0; }